Question

question given the following table of values, find h(-1) if h(x)=(x^4 + p(x))^3.

x	p(x)	p(x)
-1	-2	3
1	-3	-2
0	1	-4

provide your answer below: h(-1)=□

Explanation:

Step1: Apply the chain - rule

The chain - rule states that if $h(x)=(x^{4}+p(x))^{3}$, then $h'(x)=3(x^{4}+p(x))^{2}(4x^{3}+p'(x))$.

Step2: Substitute $x = - 1$

When $x=-1$, we first find the values of $p(-1)$ and $p'(-1)$ from the table. From the table, $p(-1)=-2$ and $p'(-1)=3$. Then substitute $x = - 1$ into $h'(x)$:
\[

$$\begin{align*} h'(-1)&=3((-1)^{4}+p(-1))^{2}(4(-1)^{3}+p'(-1))\\ &=3(1 - 2)^{2}(4(-1)+3)\\ &=3\times(-1)^{2}\times(-4 + 3)\\ &=3\times1\times(-1)\\ &=-3 \end{align*}$$

\]

Answer:

$-3$