• If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.

• Whenever you search in PBworks, Dokkio Sidebar (from the makers of PBworks) will run the same search in your Drive, Dropbox, OneDrive, Gmail, and Slack. Now you can find what you're looking for wherever it lives. Try Dokkio Sidebar for free.

View

# question04

last edited by 16 years, 11 months ago

# Question 4

## The use of a graphing calculator is NOT allowed. Suppose the ƒ(0) = 3 and ƒ' is the function shown in the graph.

Let g(x) = (x2 + 1)•ƒ(x)

(a) Evaluate g'(0).

(b) Is g increasing at x = 1? Justify.

(c) Estimate g''(0).

(d) Is g concave up at x = 1? Justify.

Solution

a. g'(0) = (x^2 + 1) * f ' (x) + 2x * f(x)

g'(0) = 1

b. g'(1) = (1^2 + 1) * f ' (1) + 2(1) * f(1)

= 2*2 + 2* f(1)

note: We are not sure about the value of f(1), BUT we can say that it is in fact a positive number since f(0) is equal to 3 and the graph of f'(x) is positive from that point onwards, therefore f(1) will definitely be greater than 3.

- So, g'(1) is increasing at x = 1 because 4 + 2 * (+no.) is equal to a positive value.

looking at the graph, I have estimated that f(0) is quite close to .9 rather than 1. So,

= .9 + 6

g(0) = 6.9

d. g''(1) = (1+1) * f(1) + 2(1) * f ' (1) + 2(1) * f '( 1) + 2 * f(1)]

note: As we can see, f(1) is hard to determine but because f'(1) is increasing, we can say that f''(1) is a positive number.

So, summing up the products, we get (if not exactly) a positive value. Therefore, g is concave up at x = 1 because its second derivative g'' is positively valued at that point.