Suppose the ƒ(0) = 3 and ƒ' is the function shown in the graph.

Let *g*(*x*) = (*x*^{2} + 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)

= (0 + 1) * 1 + 0 * 3

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.

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

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.