Question

1. use a suitable linear approximation or differential to approximate 2.001^5.

Explanation:

Step1: Define the function

Let $y = f(x)=x^{5}$, and $x = 2$, $\Delta x=0.001$.

Step2: Find the derivative

The derivative of $y = f(x)=x^{5}$ using the power - rule $(x^{n})^\prime=nx^{n - 1}$ is $y^\prime=f^\prime(x)=5x^{4}$.

Step3: Evaluate the derivative at $x = 2$

$f^\prime(2)=5\times2^{4}=5\times16 = 80$.

Step4: Use the linear - approximation formula

The linear - approximation formula is $L(x)=f(a)+f^\prime(a)(x - a)$. Here, $a = 2$ and $x=2 + 0.001$. So $f(2.001)\approx f(2)+f^\prime(2)\times(2.001 - 2)$.
We know that $f(2)=2^{5}=32$, and $f^\prime(2)\times(2.001 - 2)=80\times0.001 = 0.08$.

Step5: Calculate the approximation

$f(2.001)\approx32+0.08=32.08$.

Answer:

$32.08$