AP Calculus AB / Integration and Accumulation of Change / Question No. 39

39. The graph of the function \( g \), shown in the figure above, has horizontal tangents at \( x = 4 \) and \( x = 8 \). If \( f(x) = \int_0^{\sqrt{x}} g(t) \, dt \), what is the value of \( f'(4) \)?

1/2

3/4

3/2

Answer is: option2

1/2

Solution:

Let \( u(x) = \sqrt{x} \), then:

\( f(x) = \int_0^{u(x)} g(t)\,dt \Rightarrow f'(x) = g(u(x)) \cdot u'(x) \)

So:

\( f'(x) = g(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \)

\( f'(4) = g(2) \cdot \frac{1}{2\sqrt{4}} = 2 \cdot \frac{1}{4} = \frac{1}{2} \)

\( g(2) = 2 \) from the graph

Hence option B

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