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