Question

this is section 3.1 problem 22: for y = f(x)=x - x^3, x = 1, and δx = 0.02: δy =, and f(x)δx. round to three decimal places unless the exact answer has less decimal places. hint: follow example 2.

Explanation:

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

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

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

Substitute $x = 1$ into $f'(x)$: $f'(1)=1-3(1)^{2}=1 - 3=-2$.

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

Given $\Delta x = 0.02$ and $f'(1)=-2$, then $f'(1)\Delta x=-2\times0.02=-0.04$.

Step4: Calculate $\Delta y$

First, find $f(x+\Delta x)$ and $f(x)$.
$f(x)=x - x^{3}$, when $x = 1$, $f(1)=1-1^{3}=0$.
$x+\Delta x=1 + 0.02=1.02$, $f(1.02)=1.02-(1.02)^{3}=1.02-(1.061208)=-0.041208$.
$\Delta y=f(1.02)-f(1)=-0.041208 - 0=-0.041208\approx - 0.041$.

Answer:

$f'(x)\Delta x=-0.04$, $\Delta y\approx - 0.041$