#### Transcript of AP Calculus AB - 1985 BC 6

A) Find g(0).

g(x)=f(x )

g(0)=f(0 )

g(0)=f(0)=2

B) Find the x-coordinates of all minimum points of g. Justify your answer.

g(x)=f(x )

g'(x)=2x*f(x )

g'(0)=2(0)*f(0 ) Since f'(x)>0 for all x values, we know that 0 is a

critical point, g'(x)<0 for x<0, and g'(x)>0 for x>0.

g has a minimum at x=0

What we know:

g(0)=2

x=0 is the only critical

point

g is concave up on the x

interval (- , )

**AP Calculus AB - 1985 BC 6**

The Question and Given Information

Let f be a function that is defined and twice differentiable for all real numbers x and that has the following properties:

(i) f(0)=2

(ii) f'(x) > 0 for all x.

(iii) The graph of f is concave up for all x>0 and concave down for all x<0.

Let g be the function defined by g(x)=f(x^2).

Part A and B

Part C and D

C) Where is the graph of g concave up? Justify your answer.

g'(x)= 2x*f(x )

g''(x)= 2*f'(x ) + 4x * f''(x )

2*f'(x ) and 4x * f''(x ) are positive because

Therefore g''(x)>0 and g''(x) is always positive

g(x) is concave up for all x values

D) Using the information found in parts (a), (b), and (c), sketch a possible graph of g.

2

2

2

2

2

2

2

2

8

8

2

2

2

2

(ii) f'(0)>0 for all x

(iii) f(x) is concave up for x>0

and concave down for x<0

{

g(o)=2

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