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)
= (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 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''