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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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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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Here,  Follow normal procedure of finding maxima and minima and observing saddle point 

f(x)=3x4-16x3+24x2+37

f'(x)=12x3 -48x2+48x

as we know that to find Critical point we have to equate it to zero; 

f'(x)=0

now, we get roots i.e=0,2,2

now again f"(x)=36x2-96x+48

f''(0)=48 >0 i.e local minima at 0

f''(2)=0  i.e we need furthur investigation 

now at x=2 we got f''(2)=0 ; so on x=2 we have to investigate furthur that means we have to find f "'()

f "'()=72x-96 

f "'(2)=144-96 =48

i.e f "'(2) not equal to ZERO  so it is a saddle point (or point of inflection)

therefor   here only one local minima exist so the answer is "B"

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