Question

find $f(x)$ and $f(x)$ if $f(x) = \frac{x^2}{x - 3}$. $f(x) = \square$ $f(x) = \square$

Explanation:

Step1: Apply Quotient Rule for \( f'(x) \)

The quotient rule states that if \( f(x)=\frac{u(x)}{v(x)} \), then \( f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \). Here, \( u(x)=x^2 \), so \( u'(x)=2x \); \( v(x)=x - 3 \), so \( v'(x)=1 \).
\[
f'(x)=\frac{(2x)(x - 3)-x^2(1)}{(x - 3)^2}=\frac{2x^2-6x - x^2}{(x - 3)^2}=\frac{x^2-6x}{(x - 3)^2}
\]

Step2: Apply Quotient Rule for \( f''(x) \)

Now, for \( f''(x) \), let \( u(x)=x^2-6x \), so \( u'(x)=2x - 6 \); \( v(x)=(x - 3)^2 \), so \( v'(x)=2(x - 3) \).
\[
f''(x)=\frac{(2x - 6)(x - 3)^2-(x^2-6x)\cdot2(x - 3)}{(x - 3)^4}
\]
Factor out \( 2(x - 3) \) from the numerator:
\[
f''(x)=\frac{2(x - 3)[(x - 3)(x - 3)-(x^2-6x)]}{(x - 3)^4}=\frac{2[(x^2-6x + 9)-(x^2-6x)]}{(x - 3)^3}=\frac{2(9)}{(x - 3)^3}=\frac{18}{(x - 3)^3}
\]

Answer:

\( f'(x)=\frac{x^2 - 6x}{(x - 3)^2} \)
\( f''(x)=\frac{18}{(x - 3)^3} \)