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# AP Calculus AB - 1985 BC 6

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## Aarushi Aggarwal

on 25 October 2013

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#### 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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