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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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2 Answers

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Best answer

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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3 Comments

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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edited by
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?
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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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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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