Question

Let f be a twice-differentiable function defined on the interval -1.2 less than or equal to x less than or equal to 3.2 with f(1)=2. The graph of f', the derivative of f, is shown on the right. The graph of f' crosses the x-axis at x=-1 and x=3 and has a horizontal tangent at x=2. Let g be the function given by g(x)=e^(f(x)).

View graph here: h t t p : / / a s s e t s . o p e n s t u d y . c o m / u p d a t e s / a t t a c h m e n t s / 5 0 d 1 e f f a e 4 b 0 6 9 a b b b 7 1 0 d 3 9 - b l a d e r u n n e r 1 1 2 2 - 1 3 5 5 9 3 5 7 7 5 8 1 7 - g r a p h . p n g


1. Write an equation for the line tangent to the graph of g at x=1. 2. For -1.2 is less than or equal to x is less than or equal to 3.2, find all values of x at which g has a local maximum. Justify your answer. 3. The second derivative of g is g''(x)=x^(f(x)) [(f'(x))^2 + f''(x)]. Is g''(-1) positive, negative, or zero? Justify your answer. 4. Find the average rate of change of g', the derivative of g, over the interval [1,3].

Answer 1

1.
g(x) = e^f(x)
g'(x) = e^f(x) f'(x)
g'(1) = e^f(1) f'(1) = e^2 (-4) = -4e^2
g(1) = e^f(1) = e^2
so, you want the line through (1,e^2) with slope -4e^2:
y-e^2 = -4e^2 (x-1)

2.
g has a max where g' = 0
since e^f > 0 for all x, g'=0 when f'=0. So, g has a max/min at x = -1 or x=3
Since f''<0 at x=-1, g is a max at x = -1.

3.
e^f > 0
f'(-1) = 0
f''(-1) < 0
so, g''(-1) < 0

4.
g'(3) = e^f(3) f'(3) = 0
g'(1) = -4e^2 as above
avg change is
(g'(3)-g'(1))/(3-1) = (0- -4e^2)/2 = 2e^2

Answer 2

Two particles move along the x -axis. For 0 is less than or equal to t is less than or equal to 6, the position of particle P at time t is given by p(t)=2cos((pi/4)t), while the position of particle R at time t is given by r(t)=t^3 -6t^2 +9t+3.

1. For 0 is less than or equal to t is less than or equal to 6, find all times t during which particle R is moving to the left. 2. for 0 is less than or equal to t is less than or equal to 6, find all times t during which the two particles travel in opposite directions. 3. Find the acceleration of particles P at time t=3. Is particle P speeding up, slowing down, or doing neither at time t=3? Explain your reasoning. 4. Show that during the interval (1,3), there must be at least one instant when the particle R must have a velocity of -2.