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A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve $3x^4-16x^3+24x^2+37$ is

1. $0$
2. $1$
3. $2$
4. $3$

### 1 comment

JFI. Plot of the provided function.

Answer is $(b)$.

$f(x)=3x^4-16x^3+24x^2+37$

$f'(x)=12x^3-48x^2+48x=0$

$\implies 12x\left(x^2-4x+4\right)=0$

$\implies x\left(x-2\right)^2=0$

$\implies x=0, 2$

$f''(x)=36x^2-96x+48$

At $x=0, f''(x)=48>0$ it means that  $x =0$ is local minima.

But at $x=2, f''(x)=0$ so we can't apply second derivative test. So, we can apply first derivative test.

$f'(1) = 12, f'(3) = 36$.

So, $f'(x)$ is not changing sign on either side of $2$. So, $x=2$ is neither maxima nor minima.

So, only one extremum i.e. $x=0$.

When we can't tell about maxima/minima using second derivative test then we can also take help of third derivative test.

iff $f'''(x)\neq 0$ then there won't be any maxima/minima

iff $f'''(x)= 0$ then check for $f''''(x)$ if $+ve$ then minima , if $-ve$ then maxima.
edited
Why we cant apply second derivative test? can someone please explain... when is it possible to apply second derivative test?

Also, why don't we go for third dera'ivative test?

Since the above given link isn’t working.

Here is Wikipedia link for the general idea:

https://en.wikipedia.org/wiki/Derivative_test