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question04

Page history last edited by PBworks 14 years, 5 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)

= (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.

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