Question

what is the instantaneous rate of change at x = 2 of the function given by f(x)=\frac{x^{2}-2}{x - 1}?

Explanation:

Step1: Recall derivative formula

The instantaneous rate of change of a function $y = f(x)$ is given by its derivative $f'(x)$. Use the quotient - rule. If $f(x)=\frac{u(x)}{v(x)}$, then $f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}$. Here, $u(x)=x^{2}-2$ and $v(x)=x - 1$.

Step2: Find $u'(x)$ and $v'(x)$

Differentiate $u(x)=x^{2}-2$ and $v(x)=x - 1$ with respect to $x$. $u'(x) = 2x$ and $v'(x)=1$.

Step3: Apply quotient - rule

$f'(x)=\frac{(2x)(x - 1)-(x^{2}-2)(1)}{(x - 1)^{2}}=\frac{2x^{2}-2x-x^{2}+2}{(x - 1)^{2}}=\frac{x^{2}-2x + 2}{(x - 1)^{2}}$.

Step4: Evaluate at $x = 2$

Substitute $x = 2$ into $f'(x)$. $f'(2)=\frac{2^{2}-2\times2 + 2}{(2 - 1)^{2}}=\frac{4-4 + 2}{1}=2$.

Answer:

$2$