MHB Which AP Practice Question Choice Did You Select?

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The discussion centers around a practice AP question where the answer choices did not include the correct option. A participant struggled with understanding the functions involved and initially selected option D. However, after applying the chain rule to the function h(x) = f(g(x)), they correctly calculated h'(1) to be 15, corresponding to option B. The conversation highlights the importance of correctly applying calculus principles to solve such problems. The participant expressed gratitude for the clarification but noted the absence of a thanks button.
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this is a practice AP question but the answer is not given from the A to E selection
had a hard time with this not knowing what the functions were so just using what was given I did this (bold mine) I chose (D). It appeared that some of what was give was not needed...
 
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We are given:

$$h(x)=f\left(g(x) \right)$$

Thus, by the chain rule, we find:

$$h'(x)=f'\left(g(x) \right)\cdot g'(x)$$

So, plug in $x=1$, and what do you find?
 
MarkFL said:
We are given:

$$h(x)=f\left(g(x) \right)$$

Thus, by the chain rule, we find:

$$h'(x)=f'\left(g(x) \right)\cdot g'(x)$$

So, plug in $x=1$, and what do you find?

$$-5\cdot-3 = 15$$ which is $$(B)$$
 
Correct! (Star)
 
thanks
don't see the thanks button tho?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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