Answer to Question #153058 in Calculus for Hasan

Question

Find the relative extrema using both First and Second Partial Derivative test
fx=1+8x-3x2

Answer 1

$f(x)=1+8x-3x^2$

Differentiating w.r.t $x$ ,

$f'(x)=\frac{d}{dx}(f(x))=8-6x$

Differentiating w.r.t $x$ ,

$f''(x)=\frac{d^2f(x)}{dx}=\frac{d}{dx}(f'(x))=-6$

First Derivative test :-

$f'(x)=0 \Rightarrow 8-6x=0 \Rightarrow x= \frac{4}{3}$

$\Rightarrow f(x)$ has a critical point at $x=\frac{4}{3}$

If $x<\frac{4}{3}, f'(x)=8-6x>0$

$\Rightarrow$ $f(x)$ is increasing $\forall x<\frac{4}{3}$

If $x>\frac{4}{3}$, $f'(x)=8-6x<0$

$\Rightarrow f(x)$ is decreasing $\forall x>\frac{4}{3}$

Second Derivative test :-

$f''(\frac{4}{3})=-6<0$ , as $f''(x)=-6 \forall x\in \R$

Now, $f(4/3)=1+\frac{32}{3}-\frac{16}{3}=\frac{19}{3}$

So, from First and Second derivative test, we can conclude that:-

$f(x)$ has a relative maximum, $\frac{19}{3}$ at $x=\frac{4}{3}$ .

and, $f$ doesn't have a relative minimum.

Graph :-