Answer:
[tex]h'(-4) = 5760[/tex]
Step-by-step explanation:
We are given the function:
[tex]\displaystyle h(x) = (f\circ g)(x)[/tex]
And we want to find h'(-4) given:
[tex]\displaystyle f(x) = -6x^2 + 7[/tex]
And that the equation of the tangent line of g at x = -4 is y = 10x - 8.
Recall that (f ∘ g)(x) = f(g(x)). Hence:
[tex]\displaystyle h(x) = f(g(x))[/tex]
Find h'(x). We can take the derivative of both sides with respect to x:
[tex]\displaystyle \frac{d}{dx}\left[ h(x)\right] = \frac{d}{dx}\left[ f(g(x))\right][/tex]
By the chain rule:
[tex]\displaystyle h'(x) = f'(g(x)) \cdot g'(x)[/tex]
Therefore:
[tex]\displaystyle h'(-4) = f'(g(-4)) \cdot g'(-4)[/tex]
A) Finding f'(g(-4)):
Recall that we are given that the equation of the tangent line of g at x = -4 is:
[tex]\displaystyle y = 10x - 8[/tex]
Since this is tangent, it tells us that at x = -4, line y touches g. Therefore, y(-4) = g(-4). Find y(-4):
[tex]\displaystyle \begin{aligned} y(-4)&= 10(-4) - 8 \\ \\ &= -48\end{aligned}[/tex]
Hence, g(-4) = -48. Consequently, f'(g(-4)) = f'(-48).
Find f'(x):
[tex]\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[ -6x^2 + 7\right] \\ \\ &= -12x\end{aligned}[/tex]
So:
[tex]\displaystyle \begin{aligned} f'(-48) &= -12(-48) \\ \\ &= 576\end{aligned}[/tex]
Therefore, f'(g(-4)) = 576.
B) Finding g'(-4):
Recall that by definition, the derivative of a function at a point is the slope of the tangent line to the point.
The tangent line to g at x = -4 is given by:
[tex]y = 10 x- 8[/tex]
The slope of the line is 10. Therefore:
[tex]g'(-4) = 10[/tex]
Hence:
[tex]\displaystyle \begin{aligned}h'(-4) &= f'(g(-4)) \cdot g'(-4) \\ \\ &= (576) \cdot (10) \\ \\ &= 5760 \end{aligned}[/tex]
In conclusion:
[tex]h'(-4) = 5760[/tex]
