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Theorems On Maxima Minima And Critical Points

Learn extreme value & fermat's theorem calculus, practice to find the critical points & numbers of the function and closed interval method.

The Extreme Value Theorem

  • If \('f'\) is continuous function in interval [a, b] then it attains an absolute maximum value \(f(c)\) and absolute minimum value \(f(d)\) at some values \('c'\) and \('d'\) of \(x\) in [a, b].

  • Note that absolute maximum and absolute minimum values for different functions occur at either end points of the interval or points when tangents are horizontal.

Illustration Questions

Given is the graph of a function \('f'\) in [–1, 5], choose the correct option.

A \(f\) has absolute maximum value at \(x=3\).

B \(f\) has absolute maximum value at \(x=1\).

C \(f\) has absolute minimum value at \(x=3\).

D \(f\) has absolute maximum value at \(x=5\).

×

The highest point on the graph is (1, 5)

\(\therefore\) absolute maximum value is \(f(1)=5\)

Absolute minimum value of \(f\) does not exist as there is a discontinuity at \(x=3\).

The extreme value theorem does not hold good because of discontinuity at \(x=3\).

Given is the graph of a function \('f'\) in [–1, 5], choose the correct option.

image
A

\(f\) has absolute maximum value at \(x=3\).

.

B

\(f\) has absolute maximum value at \(x=1\).

C

\(f\) has absolute minimum value at \(x=3\).

D

\(f\) has absolute maximum value at \(x=5\).

Option B is Correct

Fermat's Theorem

  • If \(f\) has a local maxima or local minima at \(x=c\) then \(f'(c)\)\(=0\) or \(f'(c)\) does not exist.
  • If \('f'\) is a differentiable function i.e. \(f'\) exists everywhere then for local maxima or local minima at \(x=c\) we must have,\(f'(c)=0\)
  • \(f'(c)\)\(=0\Rightarrow\) does not imply there is a local maxima or local minima at \(x=c\) (Converse of Fermat's Theorem is not true).

Illustration Questions

Find the number of local maxima and number of local minima of \(f\) whose graph is shown.

A 3 local maxima and 2 local minima

B 2 local maxima and 2 local minima

C No local maxima and 3 local minima

D 3 local maxima and no local minima

×

By Fermat's Theorem, local maxima or local minima occur only at \(f'(c)\)\(=0\) point (a non differentiable point).

A, B, C, D are local extreme points.

A, C are points of local maximum value and B, D are points of local minimum value.

Find the number of local maxima and number of local minima of \(f\) whose graph is shown.

image
A

3 local maxima and 2 local minima

.

B

2 local maxima and 2 local minima

C

No local maxima and 3 local minima

D

3 local maxima and no local minima

Option B is Correct

Critical Numbers of Functions

  • A critical number of a function \('f'\) is a number \('c'\) in the domain of \('f'\) such that either \(f'(c)\)\(=0\) or \(f'(c)\) does not exist.
  • To find the critical numbers of a function:
  1. Solve \(\dfrac{dy}{dx}=0\) or \(f'\)\(=0\) in the domain.
  2. Find the points where \(f'\) does not exist.

The collection of values in steps (1) and (2) will give the list of critical points.

Illustration Questions

Find the critical numbers of the function \(f(x)=x^3-6x+9x-8\)

A \(x=2,\,5\)

B \(x=1,\,3\)

C \(x=1,\,8\)

D \(x=5,\,-7\)

×

Critical numbers are those for which either \(f'(x)\)\(=0\) or \(f'\) does not exist.

Since, \(f\) is a polynomial, it is differentiable everywhere, so only set of critical points is for which \(f'\)\(=0\).

\(f'(x)\)\(=3x^2-6×2x+9=3x^2-12x+9=0\)

\(\Rightarrow\,3(x^2-4x+3)=0\)

\(\Rightarrow\,(x-3)(x-1)=0\)

\(\Rightarrow\,x=1,\,x=3\)

\(\therefore\) Critical numbers are \(x=1,\,3\)

Find the critical numbers of the function \(f(x)=x^3-6x+9x-8\)

A

\(x=2,\,5\)

.

B

\(x=1,\,3\)

C

\(x=1,\,8\)

D

\(x=5,\,-7\)

Option B is Correct

Illustration Questions

Find the critical numbers of the function \(f(x)=x^2×(6x-7)^{1/3}\).

A \(x=1,\;2,\,\dfrac{9}{2}\)

B \(x=0,\,5,\,13\)

C \(x=-2,\,-1,\,0\)

D \(x=0,\,1,\,\dfrac{7}{6}\)

×

Critical numbers are those for which either \(f'(x)\)\(=0\) or \(f'\) does not exist.

\(f'(x)\)\(=\dfrac{d}{dx}\,x^2×(6x-7)^{1/3}\)

\(=2x×(6x-7)^{1/3}+x^2×\dfrac{1}{3}(6x-7)^{-2/3}×6\)

\(=2x(6x-7)^{1/3}+\dfrac{2x^2}{(6x-7)^{2/3}}\)

\(=\dfrac{2x(6x-7)+2x^2}{(6x-7)^{2/3}}\)

\(=\dfrac{2x(7x-7)}{(6x-7)^{2/3}}\)

\(f'(x)\)\(=0\)

\(\Rightarrow\,x=0,\,x=1\)

\(f'\) does not exist for \(x=\dfrac{7}{6}\)

\(\therefore\) Critical numbers are \(x=0,\;x=1,\;x=\dfrac{7}{6}\)

Find the critical numbers of the function \(f(x)=x^2×(6x-7)^{1/3}\).

A

\(x=1,\;2,\,\dfrac{9}{2}\)

.

B

\(x=0,\,5,\,13\)

C

\(x=-2,\,-1,\,0\)

D

\(x=0,\,1,\,\dfrac{7}{6}\)

Option D is Correct

Local Maxima and Minima as a Critical Point

If \(f\) has a local maximum or minimum at \('c'\) then \('c'\) must be a critical point

  • Note that every critical number need not be a local maxima or minima but every local maxima and minima must be a critical point.
  • Critical points which are not local maximas and minimas are often the points at which horizontal tangent to curve crosses the curve.Example \(\to\) equation \(f(x)=x^3\) at \(x=0\).

Illustration Questions

Sketch the graph of a function which is continuous in [1, 6], has no local maxima or minima but has \(x=2\;\&\;x=4\) as critical point.

A

B

C

D

×

In option 'a', \(x=2\;\&\;x=4\) are critical points as tangents are horizontal, but no maxima or minima is there. \(\therefore\) Option 'a' is correct.

Sketch the graph of a function which is continuous in [1, 6], has no local maxima or minima but has \(x=2\;\&\;x=4\) as critical point.

A image
B image
C image
D image

Option A is Correct

The Closed Interval Method

To find the absolute maximum and absolute minimum value of a function in closed interval [a, b]:

  1. Find all critical numbers of \(f\) in [a, b] and then value of \(f\) at these critical numbers.
  2. Find value of \(f\) at the end point i.e. \(f(a)\) and \(f(b)\).
  3. The largest of the value from step (1) and (2)  is the absolute maximum and smallest of the values is absolute minimum value of the function. 

Some situations which can occur are shown graphically.

Illustration Questions

Find the absolute maximum and absolute minimum values of the function \(f(x)=x^5-5x^4+5x^3+1\) in \([-1,\,2]\).

A Absolute maximum value=5, Absolute minimum value =10

B Absolute maximum value=2, Absolute minimum value =–10

C Absolute maximum value=1, Absolute minimum value =1

D Absolute maximum value= –7, Absolute minimum value =10

×

Step -1 \(\to\) Find the critical point of \('f'\) in [–1, 2]. There is no \(f'\) point where \(f'\) does not exist as \(f\) is a polynomial function.

\(f'(x)\)\(=0\)

\(\Rightarrow\,\dfrac{d}{dx}\left(x^5-5x^4+5x^3+1\right)=0\)

\(\Rightarrow\,5x^4-20x^3+15x^2=0\)

\(\Rightarrow\,5x^2\left[x^2-4x+3\right]=0\)

\(\Rightarrow\,5x^2(x-3)(x-1)=0\)

\(\Rightarrow\,x=0,\,1,\,3\)

\(\Rightarrow\,x=0,\,1\) (reject 3)

Find \(f(0)=1\)

\(f(1)=1^5-5×1^4+5×1^3+1=2\)

Step-2 \(\to\) Find \(f(a)\;\&\;f(b)\) 

\(f(-1)=(-1)^5-5×(-1)^4+5×(-1)^3+1=-10\)

\(f(2)=2^5-5×2^4+5×2^3+1\)

\(=32-80+40+1=-7\)

Step-3 \(\to\)Compare \(f(0),\,f(1),\,f(-1),\,f(2)\)

Largest \(=f(1)=2=\) maximum value

Smallest \(=f(-1)=-10\) = minimum value

 

Find the absolute maximum and absolute minimum values of the function \(f(x)=x^5-5x^4+5x^3+1\) in \([-1,\,2]\).

A

Absolute maximum value=5, Absolute minimum value =10

.

B

Absolute maximum value=2, Absolute minimum value =–10

C

Absolute maximum value=1, Absolute minimum value =1

D

Absolute maximum value= –7, Absolute minimum value =10

Option B is Correct

 Critical Numbers of a Cubic Polynomial

  • Consider   \(f(x)=ax^3+bx^2+cx+d\)

Then \(f'(x)\)\(=0\)

\(\Rightarrow\,3ax^2+2bx+c=0\)

Now, this equation can have two distinct roots or one root or no real root at all.

  • \(\therefore\,f(x)=ax^3+bx^2+cx+d\) can have two, one or no critical point.

Illustration Questions

The cubic polynomial \(f(x)=2x^3-7x^2+9x+1\) has

A 2 critical points

B 1 critical point

C No critical point

D 3 critical points

×

\(f(x)=2x^3-7x^2+9x+1\)

For critical numbers, \(f'(x)\)\(=0\)

\(\Rightarrow\,6x^2-14x+9=0\)

Now this quadratic will have two, one or no real root depending on whether \(D>0,\,D=0\;or\;D<0\)

\(D=14^2-4×6×9=196-216\)

\(=-20<0\)

\(\therefore\) no real root

\(\therefore\,f(x)\) has no critical point.

The cubic polynomial \(f(x)=2x^3-7x^2+9x+1\) has

A

2 critical points

.

B

1 critical point

C

No critical point

D

3 critical points

Option C is Correct

Identification of Critical Points on the Graph

  • A critical number of a function \(f\) is a number \(c\) in the domain of \(f\) such that either \(f'(c)\)\(=0\) or \(f'(c)\) does not exist.
  • In other words, if \(f\) has a local maximum or minimum at \(c\), then \(c\) is a critical number of \(f\).

  • In figure shown, there is a horizontal tangent when \(x=A\) and \(x=B\) and vertical tangent when \(x=0\) i.e. \(f'(x)\) does not exist at \(x=0\).
  • Therefore \(x=A,\;x=B\) and \(x=0\) are critical number.

Illustration Questions

Which of the following is not a critical point of given function.

A C

B D

C B

D E

×

There neither horizontal tangent nor vertical tangent can be drawn at point 'c'.

Hence 'c' is not a critical point on graph.

Which of the following is not a critical point of given function.

image
A

C

.

B

D

C

B

D

E

Option A is Correct

Practice Now