Question

find $f(x)$ and $f(c)$.
function: $f(x) = \frac{x^2 - 8}{x - 3}$, value of $c$: $c = 1$
$f(x) = \frac{(2x)(x - 3) - (1)(x^2 - 8)}{}$ (with a red cross)
$f(c) = 3$ (with a red cross)
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Explanation:

Step1: Apply Quotient Rule

For $f(x)=\frac{u}{v}$, $f'(x)=\frac{u'v-uv'}{v^2}$, where $u=x^2-8$, $v=x-3$.
$u'=2x$, $v'=1$, so:
$$f'(x)=\frac{(2x)(x-3)-(x^2-8)(1)}{(x-3)^2}$$

Step2: Expand numerator

Multiply terms in numerator:
$$(2x)(x-3) = 2x^2-6x, \quad -(x^2-8) = -x^2+8$$
Combine: $2x^2-6x-x^2+8 = x^2-6x+8$
So $f'(x)=\frac{x^2-6x+8}{(x-3)^2}$

Step3: Substitute $c=1$ into $f'(x)$

Plug $x=1$ into derivative:
$$f'(1)=\frac{(1)^2-6(1)+8}{(1-3)^2}=\frac{1-6+8}{(-2)^2}$$

Step4: Simplify the expression

Calculate numerator and denominator:
$$\frac{3}{4}$$

Answer:

$f'(x)=\frac{x^2-6x+8}{(x-3)^2}$
$f'(c)=\frac{3}{4}$