given that\n$f(x)=x^{10}h(x)$\n$h(-1)=5$\n$h(-1)=8$\ncalculate $f(-1)$.\nhint: use the product rule and the…

Explanation:

Step1: Apply the product rule

The product rule states that ((uv)^\prime = u^\prime v+uv^\prime). Let (u = x^{10}) and (v=h(x)). Then (u^\prime=\frac{d}{dx}(x^{10}) = 10x^{9}) (by the power rule (\frac{d}{dx}(x^n)=nx^{n - 1})) and (v^\prime=h^\prime(x)). So (f^\prime(x)=10x^{9}h(x)+x^{10}h^\prime(x)).

Step2: Substitute (x = - 1)

When (x=-1), we know that (h(-1) = 5) and (h^\prime(-1)=8). Substitute into (f^\prime(x)): (f^\prime(-1)=10(-1)^{9}h(-1)+(-1)^{10}h^\prime(-1)) Since ((-1)^{9}=-1) and ((-1)^{10}=1), we have: (f^\prime(-1)=10\times(-1)\times5 + 1\times8) (f^\prime(-1)=-50 + 8)

Answer:

(-42)