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 p 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: p p 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 p p 4 + h − 2 4 + h + 2 = lim p h!0 h 4 + h + 2 4 + h − 4 = lim p h!0 h( 4 + h + 2) h = lim p h!0 h( 4 + h + 2) 1 = lim p 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 follow- ing 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) decreas- ing? [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). 6.