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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. $3$
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JFI. Plot of the provided function. 

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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. $x=0$.

Ref: https://cims.nyu.edu/~kiryl/Calculus/Section_4.3--Derivatives_and_the_Shapes_of_Graphs/Derivatives_and_the_Shapes_of_Graphs.pdf or archive

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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.
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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?
3

Since the above given link isn’t working.

Here is Wikipedia link for the general idea:

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

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