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Math 100:106 Worksheet 4: Derivatives

Sep. 19 2019

1. (Dec. 2010 Final) If f (0) = 10 and f ′ (x) > 3 for 0 < x < 4, what is the least f (4) could possibly be? Solution: If f ′ (x) > 3 for 0 < x < 4, then this means that the function f (x) is increasing at a rate greater than 3 at every point in the interval (0, 4). What is a function which increases at a rate of exactly 3? This is a function of the form 3x + constant because its derivative is exactly 3. The statement that f ′ (x) > 3 for 0 < x < 4, then implies that f (x) grows faster than 3x + constant on [0, 4]. In order to compare these two functions we should make sure that they start at the same value, namely, 10 at x = 0. So take the constant to be 10, and let g(x) = 3x + 10. Then the function f grows faster than the function g over [0, 4] and they start at the same value at x = 0. So f (4) > g(4) = 3(4) + 10 = 22. Consequently, f (4) must be at least 22. 2. (Dec. 2006 Final) There are two distinct straight lines that pass through the point (1, −3) and are tangent to the curve y = x2 . Find equations for these two lines. Solution: The equation for the tangent line to f (x) at any point x = a, is y − f (a) = f ′ (a)(x − a).

Here f (a) = a2 and f ′ (a) = 2a. So the equation becomes y = 2a(x − a) + a2 . If the tangent line passes through the point (1, −3), then it must be true that −3 = 2a(1 − a) + a2 ,

or a2 − 2a − 3 = 0. Solving for a, we get a = 3 or a = −1. Substituting these values back into the tangent line equation, we find the equations for these two lines: y = 6x − 9 and y = −2x − 1. 3. (a) Differentiate f (x) =

5x3√ +2x+1 . x

(b) Let g(x) = Ax5/2 + 3x and suppose g ′ (4) = 1. What is A? Solution: (a) By rewriting the expression, one could use multiple methods to compute the derivative which are all perfectly fine. Here I will simplify first by writing f (x) =

5x3 + 2x + 1 √ = 5x5/2 + 2x1/2 + x−1/2 . x

x3/2 + x−1/2 − 12 x−3/2 . Then the power rule gives that f ′ (x) = 25 2 (b) By the power rule, g ′ (x) = 25 Ax3/2 + 3. As g ′ (4) = 1, we have 1 = 25(A)(43/2 ) + 3 = 1 20A + 3. Solving for A gives A = − 10 . 1

Math 100:106 Worksheet 4: Derivatives

Sep. 19 2019

4. (Dec. 2006 Final) Find the values of the constants a and b for which the function ( x2 if x ≤ 2 f (x) = ax + b if x > 2 is differentiable everywhere. Solution: Away from the point x = 2, the function f (x) is given by either ax + b or x2 both of which are differentiable. So the only point of interest is at x = 2. In order for f (x) to be differentiable at x = 2, it must be at least continuous at x = 2. In particular this means that lim f (x) = f (2). x→2+

For x > 2, f (x) = ax + b, so limx→2+ f (x) = limx→2+ (ax + b) = 2a + b, and f (2) = 22 = 4, so we get the equation 2a + b = 4. (2) must We also of course require that f ′ (2) must exist. This means that limh→0 f (2+h)−f h exist, which means that the left and right limits must exist and be equal to each other:

f (2 + h) − f (2) f (2 + h) − f (2) . = lim− h h→0 h One can compute these two limits, and get that a = 4. Notice that in this example, f (2+h)−f (2) (2) limh→0+ f (2+h)−f = limx→2− f ′ (x). = limx→2+ f ′ (x) and limh→0− h h lim

h→0+

With the two pieces of information 2a + b = 4 and a = 4, we find b = −4. 5. A rectangle is growing. At t = 0 it is a square with side length 1 metre. Its width increases at a constant rate of 2 metres per second, and its length increases at a constant rate of 5 metres per second. How fast is its area increasing at time t > 0? Solution: Let A(t) be the area of the rectangle at time t, and let L(t) denote the length and W (t) denote the width. We know that A(t) = L(t)W (t). The question is asking us for A′ (t). By the product rule, we have A′ (t) = L′ (t)W (t) + L(t)W ′ (t). We are given that L′ (t) = 5, W ′ (t) = 2, and at t = 0, L(0) = 1 = W (0). Therefore L(t) = 1 + 5t and W (t) = 1 + 2t. Substituting all the information we find that A′ (t) = 5(1 + 2t) + 2(1 + 5t) = 7 + 20t metres/second. √ √ 100 + y − 100 6. (a) Evaluate lim by interpretating it as a derivative. y→0 y x (b) (Dec. 2008 Final) Find the derivative of 3+e x. (c) For which values of x is the function f (x) = xex increasing? (Problem 6 will be postponed to next class on derivatives of exponential functions.) 2...