5.3. Exercises on the curve analysis of polynomial functions

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1 .. Exercises on the curve analysis of polynomial functions Exercise : Curve analysis Examine the following functions on symmetry, x- and y-intercepts, extrema and inflexion points. Draw their graphs including all significant points on the domain given. Instruction on how to determine extrema. Mark increasing sections with a continuous red line and decreasing sections with a dotted red line.. Find the first derivative f and its zeroes. Draw the graph of f as a red line into the same coordinate system.. Formulate a rule on how to find maxima and minima of a given function f using the sign changes of the first derivative. Instruction on how to determine inflexion points. Mark convex (increasing slope/turning left) sections with a continuous green line and concave (decreasing slope/turning right) sections with a dotted green line.. Find the second derivative f and its zeroes. Draw the graph of f as a green line into the same coordinate system.. Formulate a rule on how to find inflexion points (where the changes from convex to concave or vice versa) of a given function f using the extrema of the first derivative f or the sign changes of the second derivative f. a) f(x) = x x h) f(x) = x + x b) f(x) = x x i) f(x) = x x + 6 c) f(x) = (x ) j) f(x) = x + x d) f(x) = (x ) k) f(x) = x + x + 9 e) f(x) = (x ) l) f(x) = 0 x + 6 x 9 x (no intercepts asked!) f) f(x) = x + x m) f(x) = x x + x g) f(x) = 6 x x + x n) f(x) = 0 (x + )(x ) (polynomial long division needed!) Exercise : Common points of families of functions Draw the graphs for the t given into a single coordinate system. Determine graphically the common point P(x 0 y 0 ) of all curves and prove your conjecture using the following criterion: P(x 0 y 0 ) is a common point if for any t, t the pair f t and f t meet in P, that is: f t (x 0 ) = f t (x 0 ) = y 0. a) f t (x) = x + tx + t for t = 0, ± and ± t b) f t (x) = + for t =, 0, and x t t c) f t (x) = x + x + ( t )x + + t for t = 0, ± and ± Exercise : Envelopes of families of lines Draw the graphs for the t given into a single coordinate system. Determine graphically the envelope f of the line family and prove your conjecture using the following criterion: The curve f(x) is an envelope of the family g t, if the lines g t are tangents to f at x = t, i.e. if f(t) = g t (t) and f (t) = g t (t) for any t. a) g t (x) = tx t for t = 0, ±, ±, ±, ± and ± b) g t (x) = t x + for t =,,,,,,, and t c) g t (x) = t x + t for t =,,,,,,, and

2 Exercise : Loci of extrema and inflexion points Determine the coordinates of all extrema and inflexion points as well as the formulae of their loci. The locus of a Point (x(t) y(t)) is the curve y = y(x) which results if the parameter t is eliminated from the two equations x = x(t) and y = y(t). a) f t (x) = x(x t) with t R g) f a (x) = x x with a R + * a a b) f a (x) = x + 6x + a with a R h) f t (x) = x x tx with t R + * 6t c) f a (x) = x + ax + with a R i) f t (x) = ax x with a R* d) f a (x) = ax 8x with a R j) f b (x) = x + bx with b R + * e) f t (x) = f) f k (x) = t t x x with t R + * k) f t (x) = t x + x + with t > k x x 6kx with k R + * l) f t (x) = x t with t R x Exercise : Finding formulae of functions with given properties Determine the formulae of the functions with the following properties: a) f is a polynomial of degree and point symmetric with regard to the origin O(0 0). In O it also has the tangent y = x and it intercepts the x-axis at x =. b) f is a polynomial of degree with extremum E( ) and inflexion point W(0 ). c) f is a polynomial of degree with horizontal tangent at x = 0. The tangent line through the inflexion point W( ) has the formula y = x. 7 d) f is a polynomial of degree with x-intercept N( 0), relative maximum H( 7) and relative minimum T( ). e) f is a polynomial of degree with inflexion point W( ). It intersects the parabola g(x) = x x at x 0 = and has gradient in this intersection point. f) f is a polynomial of degree and its graph is symmetric with respect to the y-axis. It has an inflexion point W( ) and the tangent line in this point also passes through the origin O(0 0). g) f is a polynomial of degree and its graph has the inflexion point W( 0) with gradient 0. The normal line through the origin O(0 0) has the gradient 8. h) f is a polynomial of degree and its graph is point symmetric with respect to the origin O(0 0). It has a horizontal tangent 6 line through the inflexion point W( f()) with y-intercept y =. i) f t is a family of cubic parabolas with common point O(0 0) and gradient t in this point. They touch the x-axis in N(t 0). j) f t is a family of cubic parabolas which are point symmetric with respect to the origin and common point P( ). Exercise 6: Optimization problems a) The vertical line x = u moves in the domain 0 u. It intersects the x-axis in P and the parabola f(x) = (x x ) in Q. The x-intercept of the parabola with positive x-value is R. Which u results in the triangle PQR with maximal area? b) The vertical line x = u intersects f(x) = (x ) in B and g(x) = (x ) + in C. Which u in the domain u yields the maximal area for the triangle ABC with additional vertex A( )? Exercise 7: Optimization problems with respect to distance and space between two graphs The vertical line x = u can move through the whole area between the graphs of f and g. P and Q are the intersection points with f and g respectively. O(0 0) is the origin.. Which u results in the maximal distance PQ?. Which u gives the maximal area for the triangle OPQ? a) f(x) = x 6x und g(x) = x b) f(x) = x und g(x) = x(x )

3 Exercise 8: Optimization problems without given coordinates a) Find the vertices of the rectangle with maximal area between the parabola f(x) = x + and the coordinate axes. b) Find the vertices of the rectangle with maximal area between f(x) = x x 0. and the x-axis. c) Find the vertices B and C of the triangle ABC with A( ) between the parabola f(x) = x and the x-axis with maximal area and horizontal base line BC. d) Fin the vertices of the rectangle with horizontal base between f(x) = x and g(x) = x 6. Exercise 9: Optimization problem with shift a) Show that f(x) = x + x is symmetrical with respect to the axis x =. Find the triangle ABC with maximal area, horizontal base line AB and vertex C on the x-axis. b) Find the vertices of the rectangle with maximal area between the parabola f(x) = x + x and the coordinate axes. Exercise 0: Optimization problem with curve family Given is the family f t (x) = x (x t). Find the u t, so that the right angled triangle with verticesa(u 0), B(t 0) and C(u f t (u)) has maximal area. Exercise : Applications a) The cost of producing a coffee machine is. The marketing department predicts the sale of machines at the price of Furthermore they guess that each price reduction by will result in 8000 additional machines sold. Find the price with maximal gain (= sales cost) b) A rectangular pen is partitioned into two also rectangular sections and built from 00 m of fence. What dimensions should be used so that the area will be a maximum? c) A rectangular pen has an area of 00 m and is bordered at one side by a straight canal. What dimensions should be used to minimize the length of the fence? d) A rectangle with maximal area must be fitted into a triangle with base length 8 cm and height 6 cm. Hint: Show that it is sufficient to examine the special case of the rectangular triangle. e) A triangular corner has broken off a rectangular window pane during transport. Find the rectangle with maximal area which can be cut out 60 cm of the remaining form shown left. f) A wooden beam with rectangular cross section of height y and width x has a carrying strength of T = kxy with the material constant k. Find the dimensions of a rectangular beam made from a cylindrical trunk 0 cm with radius r = cm. 0 cm g) A cylinder with maximal volume is to be fitted into a ball with radius R = cm. A cylinder with height h and radius r has the volume V = π r h. h) A factory produces tin cans with a content of Litre. Find radius r and 80 cm height h so that needed area of tin sheet becomes minimal. Exercise : Newton-Raphson-method Find the first five estimates of the Newton-Raphson iteration for the given first guess x 0. a) f(x) = x x + für x 0 = b) f(x) = x x + für x 0 = c) f(x) = x x für x 0 =

4 Coordinate systems for exercise Exercise a) y 0 x Exercise b) y 0 x

5 Graphs for exercise Exercise a) y x Exercise b) y 0 x

6 Exercise : Curve analysis a) See script.. Solutions to the exercises on curve analysis of polynomial functions b) f(x) = x x = x (x + ) (x ) Symmetry: f is even with f( x) = f(x) symmetry with respect to the y-axis Intercepts N ( 0), N (0 0) (double contact point), N ( 0) Derivatives: f (x) = x x = x(x ), f (x) = x and f (x) = 6x Minima: f( ) = (y-value), f ( ) = 0 (horizontal), f ( ) = > 0 (convex) relative Min / ( ), Maximum: f(0) = 0 (y-value), f (0) = 0 (horizontal tangent line), f (0) = < 0 (concave) relative Max(0 0) Inflexion points: f( ) = 9 (y-value), f ( ) = 0 with f ( ) 0 (sign change in f ) Inf/ ( 9 ) c) f(x) = (x ) Symmetry: f( + x) = f( x) point symmetry with respect to Sy (0 ) Intercepts: S x ( 0), S y (0 ) Derivatives: f (x) = x, f (x) = x und f (x) = Inflexion points: f(0) =, f (0) = 0, f (0) = 0, f (0) = > 0 saddle point S(0 ) d) f(x) = (x ) Symmetry: f is even with f(x) = f( x) Symmetry with respect to the y-axis Intercepts: S x/ ( 0), S y (0 ) Derivatives: f (x) = x, f (x) = x and f (x) = 6x Extrema: f(0) =, (y-value), f (0) = 0 (triple zero with sign change from to +, i.e. transition from decreasing to increasing with horizontal tangent line), f (0) = 0 (double zero without sign change, f (x) 0 for all x R, i.e. convex on R), f (0) = 0 relative (and absolute) minimum T(0 ) e) f(x) = (x ) Symmetry: f( + x) = f( x) Point symmetry with respect to Sy (0 ) Intercepts: S x ( 0), S y (0 ) Derivatives: f (x) = x, f (x) = x and f (x) = x Extrema: f(0) =, (y-value), f (0) = 0 (quadruple zero without sign change, f (x) 0, i.e. increasing on R), f (0) = 0 (triple zero with sign change from to +, i.e. transition from concave to convex), f (0) = 0 inflexion point W(0 ) 6

7 f) f(x) = x + x = x(x )(x + ) Symmetry: f( x) = f(x) symmetric to origin Intercepts: S(0 0) Derivatives: f (x) = x +, f (x) = x and f (x) = Extrema (f (x) = 0 and f (x) </> 0): rel Max( ) and rel Min( ) Inflexion points (f (x) = 0 and f (x) 0): Inf(0 0) g) f(x) = 6 x x + x = 6 x(x 9x + ) Symmetry: none Intercepts: S(0 0) Derivatives: f (x) = x x + = (x )(x ), f (x) = x and f (x) = Extrema (f (x) = 0 and f (x) </> 0): rel Max( 0 ) and rel Min( 8 ) Inflexion points (f (x) = 0 and f (x) 0): Inf( ) h) f(x) = x + x = x (x ) Symmetry: none Intercepts: S x (0 0) (double) and S x ( 0) Derivatives: f (x) = x + x = x(x 8 ), f (x) = x + and f (x) = Extrema (f (x) = 0 and f (x) </> 0): rel Max( 8 6 ) Max(,6,7) and rel Min(0 0) 7 Inflexion points (f (x) = 0 and f (x) 0): Inf( ) Inf(,,6) 7 i) f(x) = x x + 6 = (x x + 7) Symmetry: f( x) = f(x) Symmetry with respect to the y-axis Intercepts: S(0 6) Derivatives: f'(x) = x x = x(x 6), f''(x) = x and f'''(x) = x Extrema: f (x) = 0 and f (x) </> 0 relative Max(0 6) and relative Min / (± 6 ) Inflexion points: f (x) = 0 and f (x) 0 Inf / (± ) j) f(x) = x + x = (x )(x ) Symmetry: f( x) = f(x) Symmetry with respect to y-axis. Intercepts: S y (0 ), S / (± 0), S / (± 0) Derivatives: f (x) = x 0x = x(x ), f (x) = x 0 = (x ), f (x) = x 6 Extrema: f (x) = 0 and f (x) </> 0 Min(0 ) and Max / (± 9 ) Max /(±,8,) Inflexion points: f (x) = 0 and f (x) 0 Inf / (± ) Inf /(±0,9 0,) 7

8 k) f(x) = x + x + 9 = (x 9)(x + ) Symmetry: f( x) = f(x) symmetry to y-axis Intercepts: S y (0 9 ) and S x/(± 0) Derivatives: f (x) = x + x = x(x )(x + ), f (x) = x + = (x ) and f (x) = 6x Extrema (f (x) = 0 and f (x) </> 0): rel Max / (± ) = Max /(± 6,) and rel Min(0 9 ) = Min(0,) Inflexion points (f (x) = 0 and f (x) 0): Inf / (± 6 6 ) Inf /(±,,7) l) f(x) = 0 x + 6 x 9 x = 0 x(x 0 x + ) Symmetry: f( x) = f(x) point symmetry with respect to origin Intercepts: S(0 0) and S x/// ( 0, 0) S x/ (±,8 0) and S x/ (±,6 0) (not asked) Derivatives: f'(x) = x + x 9 = (x 9)(x ), f''(x) = x + x and f'''(x) = x + Relative maxima (f (x) = 0 and f (x) < 0): Max ( ) = Max (,6 ) and Max ( 8 ) = Max (,6) Relative minima (f (x) = 0 and f (x) > 0): Min ( 8 ) = Min (,6) and Min ( ) = Min (,6 ) Inflexion points: f (x) = 0 and f (x) 0 Inf (0 0) and Inf / (± m) f(x) = x x + x = x(x x + 0) ) Inf / (±,,) Symmetry: f( x) = f(x) point symmetry with respect to origin Intercepts: S (0 0) Derivatives: f'(x) = (x x + ) = (x )(x ), f''(x) = x x = x(x ) und f'''(x) = x Relative maxima (f (x) = 0, f (x) < 0): Max ( ) Max (, ) and Max ( 8 ) = Max (,6 ), Relative minima (f (x) = 0, f (x) > 0): Min ( 8 ) = Min (,6 ) and Min ( ) Min (, ) Inflexion points (f (x) = 0 and f (x) 0): Inf (0 0) and Inf / (± n) f(x) = 0 (x + )(x ) = 0 (x x + 6x + x 8) Symmetry: none Intercepts: S y (0 0,8), S x ( 0), S x ( 0) (triple) 6 8 ) Inf /(±,8,) f'(x) = 0 (x x + x + ) = 0 (x + )(x ), f''(x) = (x x + ) = (x + )(x ), f'''(x) = (x ) Extrema: f (x) = 0 und f (x) > 0 Min( 0,8) Saddle point: f (x) = 0, f (x) = 0 and f (x) 0 S( 0) Inflexion point: f (x) = 0 and f (x) 0 Inf( 0,06) 8

9 Exercise : Common points of families of curves a) f t (x) = x + tx + t = (x + t) (t + ) + parabolas with vertices S( t (t + ) + ) and S g ( ) t b) f t (x) = + x t t hyperbolas with symmetry centre S(t t t ) and S g( 0) c) f t (x) = x + x + ( t )x + + t = (x + t)(x )(x t) with S g ( 0) Exercise : Envelopes of families of lines a) f(x) = x b) f(x) = x c) f(x) = x Exercise : Loci of extrema and inflexion points a) see script b) relative min T a ( 9+a) with x = a a c) relative min T a with y = x + 6 d) relative min T a a a with y = x for a 0. For a = 0 the line f 0 (x) = 8x results. e) relative max H t t t f) relative min T k (k with y = x for t 0. For t = 0 the horizontal line f0 (x) = 0 results k k + ) with y = x x + and H k ( 6k + 6) with x = and inflexion points W( k ( k) 6k( k) + with y = x 6kx + g) relative max H a a a with y = x for x > 0 and relative min T a a a with y = x for x < 0 and a > 0. For a < 0 there are no extrema. W(0 0) remains stationary h) relative max H t t t with y = x and T t (t 0) with y = 0 and inflexion points W t (t t ) with y = x. i) relative max H(0 0) and rel. min T a a 7a 7 j) relative min T b b b 6 k) relative min T t ( t t l) Extrema E t (t with y = x with y = x and inflexion points W a a 7a and inflexion points W a ( + ) with y = x + and rel. max H t( b b 6 t t with y = x. t ) with y = x, rel max for t > 0 resp. rel min for t < 0 and inflexion points W t(t with y = x + ) with y = x + and W(0 ) 9t ) with y = x 9

10 Exercise : Finding functions with given properties a) f(x) = ax + bx + cx + d with: Symmetry to O(0 0) b = 0 and d = 0 f() = 0 8a + b + c + d = 0 f'(0) = c = f(x) = x x c) f(x) = ax + bx + cx + d with: f'(0) = 0 c = 0 f() = 7a + 9b + c + d = 0 f'() = 7a + 6b + c = f''() = 0 8a + b = 0 f(x) = x x 7 e) f(x) = ax + bx + cx + d with: f() = a + b + c + d = f''() = 0 6a + b = 0 f( ) = g( ) = a + b c + d = f'( ) = a b + c = f(x) = x x x g) f(x) = ax + bx + cx + dx + e with f() = 0 6a + 8b + c + d + e = 0 f'() = 0 a + b + c + d = 0 f''() = 0 8a + b + c = 0 f(0) = 0 e = 0 f'(0) = 8 d = 8 f(x) = x 6x x 8x i) f(x) = ax + bx + cx + d with f(0) = 0 d = 0 and f'(0) = t c = t f(t) = 0: 7t a + 9t b + tc + d = 0 f'(t) = 0 7t a + 6tb + t = 0 f t (x) = x x tx 6t b) f(x) = ax + bx + cx + d with: f( ) = 8a + b c + d = f'( ) = 0 a b + c = 0 f(0) = d = f''(0) = 0 b = 0 f(x) = x x 8 d) f(x) = ax + bx + cx + d with f() = 0: 6a + 6b + b + d =0 f() = 7: 8a + b + c + d = 7 f'() = 0 a + b + c = f() = a + b + c + d = f'() = 0 a + b + c= 0 9 f(x) = x x x 8 f) f(x) = ax + bx + cx + dx + e with: Symmetry to y-axis b = 0 and d = 0 f() = a + b + c + d + e = f'() = 0 a + b + c + d = 0 f''() = 0 a + 6b + c = 0 f(x) = x x h) f(x) = ax + bx + cx + dx + ex + f with: point symmetry to O(0 0) b = d = f = 0, 6 6 f() = a + 6b + 8c + d + e + f = f'() = 0 80a + b + c + d + e= 0 f''() = 0 60a + 8b + c + d = 0 f(x) = x x x 00 j) f(x) = ax + bx + cx + d with point symmetry to O(0 0) b = d = 0 and f() = 8a + b + c + d = f a (x) = ax + ( a)x Exercise 6: Optimization problems a) Area A(u) = (f(u) 0)( u) = 8 (u 7u u + 7), A (u) = 8 (u u ), A (u) = (u 7) rel max at u = endpoints A(0) = 7 8 and A() = 6 8 abs max at u = 0. 0

11 b) parabolas with vertices S f ( ) and S g ( ) intersecting in S g ( ) (right hand endpoint) Area A(u) = g h = (u ) (f(u) g(u)) = u + 0u 8, A (u) = u + 0, A (u) = relative max at u = 9 with A( ) = endpoints: A() = 0 and A() = 0 absolute max at u =. Exercise 7: Optimization problems regarding distance and space between two graphs a) Area PQ(u) = g(u) f(u) = u + 6u, PQ (u) = u + 6, PQ (u) = relative max at u = (outside) with PQ( ) =, and endpoints PQ(0) = 0 and PQ() = absolute max at u =. Fläche A(u) = g h = (u 0) (g(u) f(u)) = u + u, A (u) = u + 6u, A (u) = 6u + 6 relative max at u = with A() = (outside) and endpoints A(0) = 0 and A() = abs max at u =. b) g(x) = x(x ) has rel max H( ) and rel min T(0 ). Intersection points at x = 0, x = und x = domains: 7 Domain 0 u : Area PQ(u) = g(u) f(u) = u u + u, PQ (u) = u 0u +, PQ (u) = 6u 0 rel max at u = 0,6 9 with PQ(0,6) 0,88 and endpoints: PQ(0) = 0 and PQ() = 0 abs max at u 0,6. Area A(u) = g h = (u 0) (g(u) f(u)) = u u + u, A (u) = u u + u, A (u) = 6u u + relative max at u = Domain u : 97 0,6 with A(0,6) 0, and endpoints A(0) = 0 and A() = 0 abs max at u 0, Area PQ(u) = f(u) g(u) = u + u u, PQ (u) = u + 0u, PQ (u) = 6u + 0 rel max at u = + 9,87 with PQ(,87) 6,06 and endpoints PQ() = 0 and PQ() = 0 abs max at u,87. Area A(u) = g h = (u 0) (f(u) g(u)) = u + u u, A (u) = u + u u, A (u) = 6u + u 97 rel max at u = +, with A(,) 9,08 and endpoints A() = 0 and A() = 0 abs max at u,. 8 6 Exercise 8: Optimization problems without given coordinates a) Rectangle must be symmetrical with respect to y-axis with width u and height f(u) Area A(u) = g h = (u ( u)) (f(u) 0) = u + 8u, A (u) = 6u + 8, A (u) = u relative max at u =, with A(,) 6,6 and endpoints A( ) = 0 and A() = 0 abs max at u,. ( b) Rectangle must be symmetrical with respect to y-axis with width u and height f(u) Area A(u) = u ( f(u)) = u + 0u + 0u, A (u) = 0u + 0u + 0, A (u) = 0u + 60u rel max at u = and endpoints:, 0 u,76 with A( ) = 0 89, (width und height = 0) and A(,76) = 0 abs max at u =. c) Area A(u) = g h = (u ( u)) ( f(u)) = u + u, A (u) = u +, A (u) = 6u relative max at u =, with A(,), with endpoints: A(0) = 0 und A() = 0 absolute max at u =. d) Intersection points (f(x) = g(x)): S / (± ) Rectangle must be symmetrical with respect to y-axis with width u and height f(u) g(u) f(u) Area: A(u) = b(u) h(u) = u [f(u) g(u)] = u [9 u ] = u + 8u relative max at u = with A( ) =. Endpoints A(0) = 0 and A() = 0 absolute max at u =

12 Exercise 9: Optimization problems with shift a) f(x + ) = x +. Area: A(u) = g h = u f( + u) = u + u, A (u) = u + and A (u) = 6u relative max at u = with A( ) =. End points: A() = 0 and A(0) = 0 abs max at u = b) Area A(u) = g h = (+u ( u)) (f(+u) 0) = u + 8u, A (u) = 6u + 8, A (u) = u rel max at u =, with A(,) 6,6 and end points A(0) = 0 and A() = 0 abs max at u,. Exercise 0: Optimization problems with families of functions Area: A t (u) = g h = (t u) (0 f t(u)) = (t u) u (u t) = u tu + t u with A t (u) = u tu + t u rel max at u = t mit At ( t t ) = and endpoints: A t(0) = A t (t) = 0 abs max at u = t Exercise : Applications a) Sales numbers A(x) = (x ) = x with x = price in gain G(x) = (x ) A(x) = x x = 8 000(x 7,x + ) relative maximum at vertex with x = 7, and A(7,) = 000 machines sold and gain G(7,) = b) Length u and width w with 00 = u + w u = 00,w Area A(w) = u w = w (00,w) =,w(w 00 ) is maximal for width w = 00 = 66,6 m and length u = 00 m with area A = 6666,6 m. c) Length u and width w with 00 = u w fence length L(u) = u + 00 u 0 and width w = with L (u) = u is minimal for length u = d) Assuming a right angled triangle with hypotenuse y = 6 u Length u and height h = 6 u Area A(u) = u h = 6u u = u(u 8) is maximal for length u = cm and height h = cm with area A = cm. Using cavalieri s principle (not needed in ist full extent) or an appropriate drawing one shows easily that shifting the vertex of the triangle parallel to the base line doesn t change the dimensions of the rectangle inside. e) Area A(u) = u (80 0,u) = 0,(u 60u) is maximal for length u = 80 cm and height 0 cm, i.e. the whole strip containing the missing triangle can be cut off! f) Carrying strength: T(x) = k x y = k x (d x ) (Pythagoras) = k d x k x with T (x) = kd kx d rel max at x = with T( d ) = kd and endpoints T(0) = 0 and T(r) = 0 abs max for x = d g) Volume: V(h) = π r h = π πr R h h (Pythagoras) = π R h π h rel max at h = and endpoints V(0) = 0 and V(R) = 0. abs. max for h = R h) Volume V = π r h, surface area A(r) = π rh + π r = V r. + π r V with A (r) = r R with V ( R) + πr rel max at r = V = with A V = V endpoints lim A(r) = lim = 0 abs min for r = V r r 0, cm and h = V = r 0,8 cm. Comparison: Radius of ball r = V 6,0 cm Exercise : Newton-Raphson-method a) x 0 =, x =, x =,7, x =,0, x =,76 b) x 0 =, x =,68, x =,07, x =,, x =,078, x 6 =,066 c) x 0 =, x =,, x =,78, x =,, x =,7