given that\n$f(x)=x^{8}h(x)$\n$h(-1)=5$\n$h(-1)=8$\n\ncalculate $f(-1)$.\n\nhint: use the product rule and…

Explanation:

Step1: Apply product - rule

The product - rule states that if $f(x)=u(x)v(x)$, then $f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Here, $u(x)=x^{8}$ and $v(x)=h(x)$. So, $f^{\prime}(x)=(x^{8})^{\prime}h(x)+x^{8}h^{\prime}(x)$.

Step2: Differentiate $u(x)$ using power - rule

The power - rule states that if $y = x^{n}$, then $y^{\prime}=nx^{n - 1}$. For $u(x)=x^{8}$, $u^{\prime}(x)=8x^{7}$. So, $f^{\prime}(x)=8x^{7}h(x)+x^{8}h^{\prime}(x)$.

Step3: Substitute $x=-1$

Substitute $x = - 1$ into $f^{\prime}(x)$: [ \begin{align*} f^{\prime}(-1)&=8(-1)^{7}h(-1)+(-1)^{8}h^{\prime}(-1)\ &=8\times(-1)\times5 + 1\times8\ &=-40 + 8\ &=-32 \end{align*} ]

Answer:

$-32$