f (x) = 2x x = 2x2 + 4x 6 x 0 = 2x 2 + 4x 6 = 2(x + 3)(x 1) x = 3 or x = 1.

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1 F16 MATH 15 Test November, 016 NAME: SOLUTIONS CRN: Use only methods from class. You must show work to receive credit. When using a theorem given in class, cite the theorem. Reminder: Calculators are not permitted. 1. (5 pts) Let f(x) = x + 4x 6 ln(x). (a) (10 pts) On what intervals, if any, is f decreasing? Note that the domain of f is (0, ). f is decreasing when f < 0. The derivative of f is Setting f (x) = 0 gives f (x) = x x = x + 4x 6 x 0 = x + 4x 6 x 0 = x + 4x 6 = (x + )(x 1) x = or x = 1. Among the x values that could be critical numbers: x =, x = 0, x = 1, only x = 1 is in the domain of f. The sign of f (x) is negative on (0, 1) and positive on (1, ), so f is decreasing on (0, 1). (b) (5 pts) What are the (x, y) coordinates, if any, of the local maxima and minima? Justify your answer. The only possible location of an extreme value is x = 1. Approach #1 (using the first derivative test): Since f changes from negative to positive at x = 1, f changes from decreasing to increasing at x = 1, so f(1) is a local minimum value. The point (1, f(1)) = (1, 5) is a local minimum. Approach # (using the second derivative test): f (x) = + 6 x. Since f (1) = 0 and f (1) > 0. The point (1, f(1)) = (1, 5) is a local minimum. (c) (7 pts) On what intervals, if any, is f concave up? f is concave up when f (x) > 0, and the second derivative of f is f (x) = + 6 x. The second derivative is positive for all x in (0, ), so f is concave up on (0, ). (d) ( pts) What are the x-coordinates, if any, of the inflection points? Justify your answer. An inflection point is a point on the curve y = f(x) where f is continuous and f changes sign. f (x) > 0 for all x in the domain of f, so there are no inflection points. 1

2 . (14 pts) Let f be a differentiable function. (a) ( pts) What would be the purpose of using Newton s method on f? The purpose of Newton s Method is to approximate the root(s) of the function f (if it has any). (b) (5 pts) What is the general formula for Newton s method? x n+1 = x n f(x n) f (x n ). (c) (7 pts) Suppose we want to use Newton s method, and we know the tangent line to f at the point ( 1, ) is y = 7x If x 1 = 1, find x. Approach 1: Newton s method idea is to use the x-intercept of the tangent line to get the approximations. We just set 7x + 10 = 0 and solve for x to get the new approximation. So x = Approach : The general formula for Newton s method is x n+1 = x n f(x n) f (x n ), so x = x 1 f(x 1) f (x 1 ). f(x 1 ) = f( 1) =, and the value of f (x 1 ) = f ( 1) is the slope of the tangent line at x 1 = 1. Therefore, f (x 1 ) = f ( 1) = 7. Substituting everything in to the formula gives, x = x 1 f(x 1) f( 1) f = ( 1) (x 1 ) f ( 1) = ( 1) 7 = 10 7.

3 e x, x 0. (1 pts) Consider the continuous function f(x) = ( πx ) cos, x > 0. Then f has critical numbers at x = 0,, 4, 6, 8,... Determine the x-values of the absolute maxima and minima, if any, of f on the interval [ 1, 4]. Since f is continuous on [ 1, 4], absolute extrema can only occur at the critical numbers or endpoints of the interval, so x = 1, x = 0, x =, and x = 4 are the possible locations of absolute extrema. ( ) ( ) π() π(4) f( 1) = 1/e, f(0) = 1, f() = cos = 1, f(4) = cos = 1 The absolute maximum value is 1 and occurs when x = 0 and x = 4. The absolute minimum value is 1 and occurs when x = 4. (16 pts) Let f be a differentiable function such that f() = 4 and f (x) for all values of x. Find the largest possible value of f(5), and justify your answer. Since f is differentiable everywhere, it is differentiable on the open interval (, 5). Also, since f is differentiable everywhere, it is continuous everywhere. In particular, f is continuous on the closed interval [, 5]. Thus, f satisfies the hypotheses of the Mean Value Theorem, so there is a number c in the open interval (, 5) such that f f(5) f() (c) = or f(5) f() = f (c)(5 ). 5 We also know that f() = 4, so f (c) = f(5) 4. Finally, since f (x) for all values of x, then f (c) and The largest possible value of f(5) is 1. f(5) 4 f(5) 1.

4 5. (16 pts) Sketch the graph of a function f satisfying the following criteria. f( ) =, f(0) = 1, f() = Asymptotes: x = 0, y = 1, y = 1 f is twice-differentiable everywhere but x = 0 Intervals (, ) (, 0) (0, ) (, ) Sign of f (x) + + Intervals (, ) (, 0) (0, ) (, ) Sign of f (x)

5 6. (17 pts) Find the point (x, y) on the curve y = x + 4 that is closest to the origin. (x, y) y = x + 4 The distance between any point and the origin is D = (x 0) + (y 0) = x + y. For simplicity, consider Since y = x + 4, D 0 = x + y. D 0 (x) = x + ( x + 4) = x + x + 4, and the domain of this function is [, ). Taking the derivative gives D 0(x) = x +, we can see that the only critical number is x = 1. D 0 shows that D 0 only decreases when x < 1 and only increases when x > 1. Therefore, x = 1 is the location of a local and absolute minimum and y = ( 1) + 4 =. The point on the curve closest to the origin is (x, y) = ( 1, ). 5