MA4001 Engineering Mathematics 1 Lecture 15 Mean Value Theorem Increasing and Decreasing Functions Higher Order Derivatives Implicit Differentiation

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1 MA4001 Engineering Mathematics 1 Lecture 15 Mean Value Theorem Increasing and Decreasing Functions Higher Order Derivatives Implicit Differentiation Dr. Sarah Mitchell Autumn 2014

2 Rolle s Theorem Theorem If: f(x) is continuous in [a, b]; f(x) is differentiable in (a, b); f(a) = f(b) then, there exists a point c such that f (c) = 0.

3 Proof. f(a) = f(b) could mean that f(x) is constant in [a, b], in which case f = 0 everywhere in [a, b]. Otherwise, if f(x) is not constant, by the max-min theorem, f(x) achieves a maximum and minimum in [a, b]. At least one maximum or minimum must be at an interior point which we can call c. Thus by the previous theorem, f (c) = 0.

4 Mean Value Theorem Theorem If f(x) is continuous in [a, b]; differentiable in (a, b); then there exists a point c (a, b) such that f (c) = f(b) f(a) b a

5 Note that here: f (c) is the slope of the tangent at c. f(b) f(a) b a (b, f(b)) is the slope of the secant joining (a, f(a)) and The equation of the secant is y f(a) = f(b) f(a) (x a). b a

6 Proof of Mean Value Theorem Proof. [ Let g(x) = f(x) f(a)+ f(b) f(a) ] (x a). b a Then g(x) satisfies the conditions for Rolle s theorem: g(x) is continuous in [a, b] and differentiable in (a, b); g(a) = g(b) = 0. Thus there is a point c such that g (c) = 0. g (c) = 0 = f (c) f(b) f(a). b a Thus f (c) = f(b) f(a) b a

7 Example Prove that sin x < x for all x > 0.

8 Proof. The result is obvious for x > 1 as sin x 1 < x. For 0 < x < 1, consider f(t) = sin t on [0, x]. By the mean value theorem there exists a point c such that 0 < c < x and f (c) = f(x) f(0) x 0 = f(x) x = sin x x That is, 1 > cos c = sin x x Thus sin x < x. as 0 < c < 1

9 Increasing functions Definition Suppose f(x) is defined on an interval I. Then if, for all x 2 > x 1 I, f(x 2 ) > f(x 1 ), f is said to be increasing on I.

10 Decreasing functions Definition Suppose f(x) is defined on an interval I. Then if, for all x 2 > x 1 I, f(x 2 ) < f(x 1 ), f is said to be decreasing on I.

11 Non-decreasing functions Definition Suppose f(x) is defined on an interval I. Then if, for all x 2 > x 1 I, f(x 2 ) f(x 1 ), f is said to be non-decreasing on I.

12 Non-increasing functions Definition Suppose f(x) is defined on an interval I. Then if, for all x 2 > x 1 I, f(x 2 ) f(x 1 ), f is said to be non-increasing on I.

13 Theorem If for all x (a, b): f (x) > 0, then f is increasing in (a, b); f (x) < 0, then f is decreasing in (a, b); f (x) 0, then f is non-decreasing in (a, b); f (x) 0, then f is non-increasing in (a, b).

14 Remark f (x) > 0 means that tangent lines have positive slopes. f (x) < 0 means that tangent lines have negative slopes.

15 Proof for f (x) > 0 case For any x 1, x 2, such that a < x 1 < x 2 < b, apply the mean value theorem: There exists a point c (x 1, x 2 ), such that f(x 2 ) f(x 1 ) x 2 x 1 = f (c) > 0 Since x 2 x 1 > 0, f(x 2 ) f(x 1 ) must also be > 0. Thus f(x) is increasing. The other 3 cases can be proved analogously.

16 Example: On which intervals is f(x) = x 3 12x + 1 increasing or decreasing? f (x) = 3x 2 12 = 2(x + 2)(x 2) f (x) > 0, i.e., f(x) is increasing if x > 2 or x < 2. f (x) < 0, i.e., f(x) is decreasing if 2 < x < 2.

17 Higher order derivatives The derivative of y = f (x) is called the second derivative of f : denoted in various ways as f (x) = (f (x)) y = f (x) = d 2 y dx 2 = d dx Similarly the n-th derivative of f is f (n) (x) = i.e., f(x) differentiated n times. d dx f(x) = d 2 dx 2 f(x) (...( (f ) )... ) (x) Note that the following notations can also be used, in particular for higher derivatives: f (0) (x) = f(x), f (1) (x) = f (x), f (2) (x) = f (x), f (3) (x) = f (x),...

18 Example: nth degree polynomial p(x) = x n + a n 1 x n 1 + a n 2 x n 2 + +a 3 x 3 + a 2 x 2 + a 1 x + a 0 p (x) = nx n 1 +(n 1)a n 1 x n 2 +(n 2)a n 2 x n a 3 x 2 +2a 2 x+a 1 p (x) = n(n 1)x n 2 +(n 1)(n 2)a n 1 x n 3 + (n 2)(n 3)a n 2 x n a 3 x + 2a 2 p (x) = n(n 1)(n 2)x n 3 +(n 1)(n 2)(n 3)a n 1 x n 4 + (n 2)(n 3)(n 4)a n 2 x n a 3. Differentiating n times we obtain p (n) = n(n 1)(n 2)(n 3) = n!

19 Example: y = 1 x y = x 1 y = ( 1)x 2 y = ( 1)( 2)x 3 y = ( 1)( 2)( 3)x 4. y (n) = ( 1)( 2)( 3)...( n)x (n+1) = ( 1)n n! x n+1

20 Example: differential equation If y = A cos(kx)+bsin(kx), then y = ka sin(kx)+kb cos(kx) y = k 2 A cos(kx) k 2 B sin(kx) Therefore y satisfies the differential equation y + k 2 y = 0. This is the differential equation of simple harmonic motion. For example, it describes the motion of a mass suspended from a fixed base by a spring (with no damping).

21 Implicit differentiation A function may be defined: explicitly: y = f(x) e.g., y = x 3, y = x, or implicitly: i.e., through an equation F(x, y) = 0. e.g., x 2 + y 2 = 2 2.

22 Often we cannot solve F(x, y) = 0 to obtain an explicit representation for y, but the derivative y may still be defined. We differentiate F(x, y) = 0 with respect to x, regarding y as a function of x and using the chain rule.

23 Example: Find dy dx where y sin x = x 3 + cos y. We differentiate both sides of the equation with respect to x: d dx (y sin x) = d dx (x 3 )+ d (cos y) dx y cos x + dy dx sin x = 3x 2 sin y dy dx by the product rule by the chain rule dy dx (sin x + sin y) = 3x 2 y cos x dy dx = 3x 2 y cos x sin x + sin y