Question

this is section 3.1 problem 16: for y = f(x)=\frac{3}{x^{2}}, x = 1, and delta x=0.03: delta y=square, and f(x)delta x=square. round to three decimal places unless the exact answer has less decimal places. hint: follow example 2. resources ebook

Explanation:

Step1: Find the derivative of $y = f(x)=\frac{3}{x^{2}}=3x^{- 2}$.

Using the power - rule for differentiation $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f'(x)=3\times(-2)x^{-2 - 1}=-6x^{-3}=-\frac{6}{x^{3}}$.

Step2: Evaluate $f'(x)$ at $x = 1$.

Substitute $x = 1$ into $f'(x)$, then $f'(1)=-\frac{6}{1^{3}}=-6$.

Step3: Calculate $f'(x)\Delta x$.

Given $\Delta x = 0.03$ and $f'(1)=-6$, then $f'(x)\Delta x=f'(1)\Delta x=-6\times0.03=-0.180$.

Step4: Calculate $\Delta y$.

First, find $f(x+\Delta x)$ and $f(x)$. When $x = 1$ and $\Delta x=0.03$, $x+\Delta x = 1.03$.
$f(1)=\frac{3}{1^{2}} = 3$, $f(1.03)=\frac{3}{(1.03)^{2}}=\frac{3}{1.0609}\approx2.828$.
$\Delta y=f(1.03)-f(1)\approx2.828 - 3=-0.172$.

Answer:

$f'(x)\Delta x=-0.180$, $\Delta y=-0.172$