Math 220 GW 7 SOLUTIONS 1. Using the Limit Definition of Derivative, Find the Derivative Function, F (X), of the Following Funct

Math 220 GW 7 SOLUTIONS

1. Using the limit deﬁnition of derivative, ﬁnd the derivative function, f 0(x), of the following functions. Show all your beautiful algebra.

(a) f(x) = 2x

f(x + h) − f(x) 2(x + h) − 2x lim = lim h→0 h h→0 h 2x + 2h − 2x = lim h→0 h 2h = lim h→0 h 2.

(b) f(x) = −x2 + 2x

f(x + h) − f(x) −(x + h)2 + 2(x + h) + x2 − 2x lim = lim h→0 h h→0 h −x2 − 2xh − h2 + 2x + 2h + x2 − 2x = lim h→0 h −2xh − h2 + 2h = lim h→0 h = lim(−2x − h + 2) h→0 = −2x + 2.

1 2. You are told f(x) = 2x3 − 4x, and f 0(x) = 6x2 − 4. Find f 0(3) and f 0(−1) and explain, in words, how to interpret these numbers.

f 0(3) = 6(3)2 − 4 = 50. f 0(1) = 6(1)2 − 4 = 2. These are the slopes of f(x) at x = 3 and x = 1. Both are positive, thus f is increasing at those points. Also, 50 > 2, so f is increasing faster at x = 3 than at x = 1. Example Find the derivative of f(x) = 3/x2.

f(x + h) − f(x) f 0(x) = lim h→0 h 3 3 2 − 2 = lim (x+h) x h→0 h x2 3 3 (x+h)2 2 ∗ 2 − 2 ∗ 2 = lim x (x+h) x (x+h) h→0 h 3x2−3(x+h)2 2 2 = lim x (x+h) h→0 h 1 3x2 − 3(x + h)2 1 = lim ∗ h→0 x2(x + h)2 h 3x2 − 3(x2 + 2xh + h2) = lim h→0 hx2(x + h)2 3x2 − 3x2 − 6xh + h2 = lim h→0 hx2(x + h)2 h(−6x + h) = lim h→0 hx2(x + h)2 h −6x + h = lim ∗ h→0 h x2(x + h)2 −6x + h = lim h→0 x2(x + h)2 −6x + 0 = x2(x + 0)2 −6x = x4 −6 = x3 3. For the following questions consider f(x) = 4/x.

2 (a) Find the derivative f 0(x).

f(x + h) − f(x) 4 − 4 lim = lim x+h x h→0 h h→0 h 4x−4(x+h) = lim x(x+h) h→0 h 4h = lim h→0 hx(x + h) 4 = lim h→0 x(x + h) 4 = . x2

(b) Find and interpret f 0(5). 4 4 f 0(5) = = . 52 25 4 The slope of f at x = 5 is 25 . (c) When x = 5, is the graph of f(x) increasing, decreasing, or nei- ther? Explain why. 4 Since 25 is positive, f(x) is increasing at x = 5.

3 √ 4. Let f(x) = x − 5

(a) Find the equation of the secant line that goes through the graph when x = 9 and x = 14. First, we ﬁnd the slope of the secant line: √ √ f(14) − f(9) 14 − 5 − 9 − 5 3 − 2 1 = = = . 14 − 9 5 5 5 Then, we use point-slope formula for a line using the point (9, f(9)) = (9, 2): 1 y − 2 = (x − 9). 5

(b) Find the equation of the tangent line to the graph of f(x) when f(9 + h) − f(9) x = 9.(Hint: Calculate lim to ﬁnd the slope of h→0 h the tangent line at x = 9) Finding the tangent line, we need the tangent slope:

f(9 + h) − f(9) p(9 + h) − 5 − 2 lim = lim h→0 h h→0 h √ √ 4 + h − 2 4 + h + 2 = lim √ h→0 h 4 + h + 2 4 + h − 4 = lim √ h→0 h( 4 + h + 2) h = lim √ h→0 h( 4 + h + 2) 1 = lim √ h→0 4 + h + 2 1 = . 4 Using point-slope formula using the point (9, f(9)) = (9, 2): 1 y − 2 = (x − 9). 4

4 5. Match the 3 functions below to their derivatives. Do this by consid- ering when a function is increasing its derivative is positive, when a function is decreasing its derivative is negative, and when a function has a min/max its derivative is 0.

Functions

Possible Derivatives

Function 1—Derivate 2; Function 2—Derivate 1; Function 3—Derivate 3.

5 6. Consider the graph below to be the graph of f 0(x), the DERIVATIVE of some unknown function f(x). Use this graph to answer the following questions.

(a) What is the slope of the tangent line to the graph of f(x) when x = 1? f 0(1) = 1. (b) Knowing that the function f(x) can only have a min/max at the point x = a if f 0(a) = 0, what are the possible mins/maxes of f(x)? x = 1.6, 3.2, 4.7. (c) Knowing the the function f(x) can only be increasing at the point x = a if f 0(a) is positive, on what intervals is f(x) increasing? [−1, 1.6] ∪ [3.2, 4.7]. Note: we don’t know if it extends below −1. (d) Knowing that the function f(x) can only be decreasing at the point x = a if f 0(a) is negative, on what intervals is f(x) decreasing? [1.6, 3.2] ∪ [4.7, 5]. Note: we don’t know if it extends past 5 (e) Find all points x = a where the tangent line to f(x) is horizontal. These are the same as part (b): x = 1.6, 3.2, 4.7. (f) Sketch a graph of f(x).