Informative line

Local Maxima And Minima

Practice first and second derivative test for local extrema, find local maxima and local minima of a function by second derivative test & concavity test.

Concavity of a Curve

• Concavity describes the nature of bend in a curve.
• Concavity can be upwards or downwards.
• This can be identified visually.

Identification of concavity

• If the graph of function  $$'f'$$  lies above all the tangents in an interval then it is called concave upwards on that interval.

• If the graph of function $$'f'$$ lies below all its tangents  in an  interval , it is called concave downwards in that interval.

For example:

Consider the curve as shown in the figure.

We will identify the concavity of the curve in the interval $$(a,\;b)$$ .

Step 1: Draw tangents to the curve $$(a,\;b)$$ at any points on the curve.

We draw tangent at $$P_1,\;P_2\;\text{and}\;P_3$$

Step 2: Visualize.

Here, we can see that the curve lies below the tangent at any given point of contact.

So, we can any say that the curve is concave downward.

Here, we can see that the curve lies below the tangent at any given point of contact.

So, we can any say that the curve is concave downward.

Which of the following curve is concave upwards in the interval (a,b)?

A

B

C

D

×

For option (A),

Draw tangents to the curve (a,b) at any points  P1, P2, P3

Here we can see that the curve lies below the tangent at any given point of contact. So we can say that the curve is concave downwards.

Thus, this option is incorrect.

For option (B),

Draw tangents to the curve (a,b) at any  points  P1, P2, P3

Here we can see that the curve lies below the tangent at any given point of contact. So we can say that the curve is concave downwards.

Thus, this option is incorrect.

For option (C),

Draw tangents to the curve (a,b) at any points  P1, P2, P3

Now  we can see that the curve lies above the tangent at any given point of contact .So we can say that the curve is concave upwards  in the  interval (a,b).

Thus, this option is correct.

For option (D),

Draw tangents to the curve (a,b) at any points  P1, P2, P3

Here, we can see that the curve lies below the tangent at any given point of contact. So, we can say the curve is concave downwards.

Thus, the option is incorrect.

Which of the following curve is concave upwards in the interval (a,b)?

A
B
C
D

Option C is Correct

Identification of a Critical Point (whether it is a Local Maxima or a Local Minima)

Critical points

Critical points are the points in an interval at which the function may obtain a maximum or minimum value.

First derivative test :

Let C be a critical number of a continuous function $$f$$

• If $$f'$$ changes sign from negative to positive at C(going from right to left) then $$f$$ has a local maxima at C.

• If $$f'$$ changes sign from positive to negative(going from right to left) at C then $$f$$ has a local minima at C.

• If $$f'$$ does not change sign at 'C' (it is positive or negative on either side of C) then $$f$$ has no local maxima or local minima at C.

For example:

We will find out the local maximum and local minimum values of the function $$f(x)=2x^3-6x$$ by first derivative test.

Step 1: Find first derivative of function $$f(x)$$

$$f'(x)=\dfrac{d}{dx}(2x^3-6x)$$

$$f'(x)=6x^2-6$$

Step 2: Calculate critical points by putting $$f'(x)=0$$

$$f'(x)=0$$

$$6x^2-6=0$$

$$6(x^2-1)=0$$

$$6(x+1)(x-1)=0$$

$$x=1,\;-1$$

So, $$f$$ has a maxima at $$x=-1$$ and a minima at $$x=1$$

Thus,

$$f(1)=-4=$$ local minimum value of $$f$$

$$f(-1)=4=$$ local maximum value of $$f$$

Find the local maximum and local minimum value of $$f(x) = x^3 - 6x^2+9x+15$$ by first derivative test .

A Local maximum value = $$f(2)=17$$, Local minimum value = $$f(-1) =-1$$

B Local maximum value = $$f(1)=19$$, Local minimum value = $$f(3) =15$$

C Local maximum value = $$f(0)=15$$, Local minimum value = $$f(3) =15$$

D Local maximum value = $$f(5)=35$$, Local minimum value = $$f(-1) =-1$$

×

To find local maxima or local minima  first calculate critical numbers, for critical numbers , $$f'(x)= 0$$

$$f(x) = x^3-6x^2+9x+15$$

$$\Rightarrow f'(x) = 3x^2-12\,x+9=0$$

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

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

$$\Rightarrow x=3$$ or  $$x =1$$

$$f$$ has a maxima at $$x=1$$ and a minima at $$x=3$$

$$f(1) =19=$$ local maximum value of $$f$$

$$f(3) =15=$$ local minimum value of $$f$$

Find the local maximum and local minimum value of $$f(x) = x^3 - 6x^2+9x+15$$ by first derivative test .

A

Local maximum value = $$f(2)=17$$, Local minimum value = $$f(-1) =-1$$

.

B

Local maximum value = $$f(1)=19$$, Local minimum value = $$f(3) =15$$

C

Local maximum value = $$f(0)=15$$, Local minimum value = $$f(3) =15$$

D

Local maximum value = $$f(5)=35$$, Local minimum value = $$f(-1) =-1$$

Option B is Correct

Find the local maximum and local minimum value of the function  $$f(x) = \dfrac{x^2+x+1}{x^2-x+1}$$ by first derivative test .

A Local maximum value = $$f(0)=1$$, Local minimum value = $$f(-1) =\dfrac{1}{3}$$

B Local maximum value = $$f(1)=3$$, Local minimum value = $$f(-1) =\dfrac{1}{3}$$

C Local maximum value = $$f(2)=\dfrac{7}{3}$$, Local minimum value = $$f(-2) =\dfrac{3}{7}$$

D Local maximum value = $$f(4)=\dfrac{21}{13}$$, Local minimum value = $$f(0) =1$$

×

To find local maxima or local minima  first calculate critical numbers.

For critical numbers ,

$$f'(x)= 0$$

$$f'(x) = \dfrac{x^2+x+1}{x^2-x+1} \Rightarrow f'(x) = \dfrac{(x^2-x+1)(2x+1)-(x^2+x+1)(2x-1)}{(x^2-x+1)^2}$$

$$\therefore f'(x)=0 \Rightarrow \dfrac{2x^3+x^2-2x^2-x+2x+1-2x^3+x^2-2x^2+x-2x+1}{(x^2-x+1)^2}=0$$

$$\Rightarrow \dfrac{-2x^2+2}{(x^2-x+1)^2}=0 \Rightarrow \dfrac{2(1-x)(1+x)}{(x^2-x+1)^2}=0$$

$$\Rightarrow x=1 \,\,\,\,\,or \,\,\,\,x=-1$$

$$f$$  has local maxima at $$x =1$$ and local minima at $$x =-1$$

$$f(1) =3=$$ local maximum value

$$f(-1) =\dfrac{1}{3}=$$ local minimum value

Find the local maximum and local minimum value of the function  $$f(x) = \dfrac{x^2+x+1}{x^2-x+1}$$ by first derivative test .

A

Local maximum value = $$f(0)=1$$, Local minimum value = $$f(-1) =\dfrac{1}{3}$$

.

B

Local maximum value = $$f(1)=3$$, Local minimum value = $$f(-1) =\dfrac{1}{3}$$

C

Local maximum value = $$f(2)=\dfrac{7}{3}$$, Local minimum value = $$f(-2) =\dfrac{3}{7}$$

D

Local maximum value = $$f(4)=\dfrac{21}{13}$$, Local minimum value = $$f(0) =1$$

Option B is Correct

Finding the Values of Parameters (when the Local Maxima and Minima Values of a Function are given)

If a function contains some parameters  other than independent variable  $$x$$  , and we are given the values of $$x$$  at which local maxima or minima occurs then

1. Find $$f'(x) \to$$ it will contain the parameter which is to be evaluated.
2. Put  $$f'(x) =0$$  at those $$x$$ where maxima and minima  are given , this will result in a equation in the parameters.
3. Solve the equation.

Note :  That there can be more than one parameter in same cases.

• Consider an example to understand this.

we have to find the value of $$'a'$$ in $$f(x)=2x^2+ax+3$$, which has a local maximum value of $$17$$ at $$x=2$$

For local maximum or minimum values, $$f'(x)=0$$

Step 1: Find first derivative of function.

$$f'(x)=\dfrac{d}{dx}(2x^2+ax+3)$$

$$f'(x)=4x+a$$

Step 2: put $$f'(x)=0$$ and solve for $$a$$

$$f'(x)=0$$

$$4x+a=0$$

As $$f'(x)$$ should be zero at $$x=2$$  (slope is zero at point of local maxima and minima) so, $$4x+a$$ should be zero at $$x=2$$,

So,

$$4(2)+a=0$$

$$a=-8$$

So, $$\boxed{a=-8}$$

Find a function $$f(x) = a\,x^3+b\,x^2+c\,x+d$$  which has local maximum value of 19 at $$x =1$$ , and local minimum value  of 15 at $$x =3$$.

A $$f(x)= 2\,x^3 + 3\,x^2 -x+8$$

B $$f(x)= \,x^3 + 2\,x^2 -x+7$$

C $$f(x)= 3\,x^3 + 5\,x^2 +x+1$$

D $$f(x)= \,x^3 - 6\,x^2 +9\,x+15$$

×

For local maximum  or minimum value $$f' (x)=0$$

$$f' (x)=0$$

$$\Rightarrow 3\,a\,x^2+2\,b\,x+c=0$$

The roots of this equitation  should be $$x = 1\,\,{\text &}\,\,x=3$$ .

$$\therefore \,3\,a+2\,b+c=0$$  .....(1)

and  $$27\,a+6\,b+c=0$$  ......(2)

Also $$f(1) =19$$

$$\Rightarrow a+b+c+d =19$$      .......(3)

$$f(3) =15$$

$$\Rightarrow 27\,a+9\,b+3\,c+d =15$$    ........(4)

Solving (1), (2), (3) and (4) equation we get

$$\Rightarrow a=1,\,b=-6,\,c=9,\,d=15$$

$$\therefore \,f(x) = x^3 -6\,x^2+9\,x+15$$

Find a function $$f(x) = a\,x^3+b\,x^2+c\,x+d$$  which has local maximum value of 19 at $$x =1$$ , and local minimum value  of 15 at $$x =3$$.

A

$$f(x)= 2\,x^3 + 3\,x^2 -x+8$$

.

B

$$f(x)= \,x^3 + 2\,x^2 -x+7$$

C

$$f(x)= 3\,x^3 + 5\,x^2 +x+1$$

D

$$f(x)= \,x^3 - 6\,x^2 +9\,x+15$$

Option D is Correct

Concavity Test

• Concavity of a curve can also be identified by using second derivative of the function of that curve.
• In this test, we will first find out the second derivative of the function.
• There may be either of the following two cases.

Case-1 If $$f''(x)>0$$ for all $$x$$ in an interval, then graph of $$f$$ is concave upwards in that interval.

Case-2 If $$f''(x)<0$$ for all $$x$$ in an interval, then graph of $$f$$ is concave downwards in that interval.

Finding interval using concavity test

(i) When curve is concave upwards

To find the interval in which the curve is concave upwards, solve $$f''(x)>0$$.

(ii) When curve is concave downwards

To find the interval in which the curve is concave downwards, solve $$f''(x)<0$$ .

For example: We will find the intervals of concavity of the function $$f(x)=x^4-6x^3+12x^2-10$$

For concave upwards : $$f''(x)>0$$

For concave downwards : $$f''(x)<0$$

Step 1: First calculate $$f''(x)$$

$$f'(x)=4x^3-18x^2+24x$$

$$f''(x)=12x^2-36x+24$$

Step 2: Put $$f''(x)=0$$

$$12x^2-36x+24=0$$

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

$$x^2-3x+2=0$$

$$x^2-x-2x+2=0$$

$$x(x-1)-2(x-1)=0$$

$$(x-1)(x-2)=0$$

Step 3: We will make a table for the sign

Interval (x–1) (x–2) f''
x>2 + + +
1<x<2 +
x<1 +

From the table, we have

$$f$$ is concave upwards on $$(-\infty,\;1)\;\cup(2,\;\infty)$$

and concave downwards on $$(1,\;2)$$

Find the  intervals of concavity of the function $$f(x) = x^4-12\,x^3+48\,x^2-50$$.

A Concave upwards on $$(-\infty, 4) \cup (6,\infty)$$ , Concave  downwards on $$(4,6)$$

B Concave upwards on $$(-\infty, 2) \cup (4,\infty)$$ , Concave  downwards on $$(2,4)$$

C Concave upwards on $$(-\infty, 1) \cup (5,\infty)$$ , Concave  downwards on $$(1,5)$$

D Concave upwards on $$(-\infty, 0) \cup (2,\infty)$$ , Concave  downwards on $$(0,2)$$

×

For concave upwards $$\to f''(x) >0$$

For  concave downwards $$\to f'' (x) <0$$

$$\Rightarrow f'(x)= 4x^3 - 36x^2+96x$$

$$\Rightarrow f''(x)= 12x^2 - 72x+96 = 12(x^2-6x+8)$$

$$= 12(x-4)(x-2)$$

 Interval $$(x-4)$$ $$(x-2)$$ $$f''$$ $$x>4$$ $$+$$ $$+$$ $$+$$ $$2 From the table \(f$$ is concave upwards on $$(-\infty, 2) \cup (4,\infty)$$ and concave  downwards on $$(2,4)$$

Find the  intervals of concavity of the function $$f(x) = x^4-12\,x^3+48\,x^2-50$$.

A

Concave upwards on $$(-\infty, 4) \cup (6,\infty)$$ , Concave  downwards on $$(4,6)$$

.

B

Concave upwards on $$(-\infty, 2) \cup (4,\infty)$$ , Concave  downwards on $$(2,4)$$

C

Concave upwards on $$(-\infty, 1) \cup (5,\infty)$$ , Concave  downwards on $$(1,5)$$

D

Concave upwards on $$(-\infty, 0) \cup (2,\infty)$$ , Concave  downwards on $$(0,2)$$

Option B is Correct

Second Derivative Test

• Using second derivative test, we can find the local maximum and local minimum values within an interval.
• For this, we will calculate the second derivative of the given function.
• There may be either of the following two cases.
• If $$f''$$ is continuous near 'C', and $$f'(C) =0 \,\,\,{\text &} \,\,f'' (C) <0\Rightarrow f$$ has a local maxima at $$x=C$$

• If $$f''$$ is continuous near 'C', and $$f'(C) =0 \,\,\,{\text &} \,\,f'' (C) >0\Rightarrow f$$ has a local minima at $$x=C$$.

For example:

We will find local maximum and local minimum values of $$f(x)=\cos x(1+\sin x)$$ by second derivative test $$(0<x<2\pi)$$.

Step 1: Calculate $$f'(x)$$

$$f'(x)=\dfrac{d}{dx}(\cos x)+\dfrac{d}{dx}(\cos x\,\sin x)$$

$$f'(x)=-\sin x+\cos x\cdot\cos x-\sin x\;\sin x$$

$$f'(x)=-\sin x+\cos^2x-\sin ^2x$$

$$f'(x)=-\sin x+1-\sin^2x-\sin^2x$$

$$f'(x)=-2\sin ^2x-\sin x+1$$

Step 2: Put $$f'(x)=0$$ and find values (critical points) for $$x.$$

$$f'(x)=0$$

$$-2\sin^2x-\sin x+1=0$$

$$(\sin x+1)(2\sin x-1)=0$$

$$\sin x=-1,\;+\dfrac{1}{2}$$

$$x=\dfrac{\pi}{6},\;\dfrac{5\pi}{6},\;\dfrac{3\pi}{2}$$

These are the critical points.

Step 3: Find $$f''(x)$$.

$$f'(x)=-2\sin^2x-\sin x+1$$

$$f''(x)=\dfrac{d}{dx}(-2\sin^2x)-\dfrac{d}{dx}(\sin x)+\dfrac{d}{dx}(1)$$

$$f''(x)=-4\sin x\;\cos x-\cos x+0$$

$$f''(x)=-4\sin x\;\cos x-\cos x$$

Step 4: Find the values of $$f''(x)$$ at critical points.

$$f''\left(\dfrac{\pi}{6}\right)=-4\sin\left(\dfrac{\pi}{6}\right)\cos\left(\dfrac{\pi}{6}\right)-\cos\left(\dfrac{\pi}{6}\right)$$

$$f''\left(\dfrac{\pi}{6}\right)=-4\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt 3}{2}\right)-\left(\dfrac{\sqrt3}{2}\right)$$

$$=-\sqrt3-\dfrac{\sqrt3}{2}<0$$

$$f''\left(\dfrac{5\pi}{6}\right)=-4\sin\left(\dfrac{5\pi}{6}\right)\cos\left(\dfrac{5\pi}{6}\right)-\cos\left(\dfrac{5\pi}{6}\right)$$

$$=-4\left(\dfrac{1}{2}\right)\left(\dfrac{-\sqrt3}{2}\right)-\left(\dfrac{-\sqrt3}{2}\right)$$

$$=\sqrt3+\dfrac{\sqrt3}{2}>0$$

$$f''\left(\dfrac{3\pi}{2}\right)=-4\sin\left(\dfrac{3\pi}{2}\right)\cos\left(\dfrac{3\pi}{2}\right)-\cos\left(\dfrac{3\pi}{2}\right)$$

$$=-4(-1)(0)-(0)$$

$$=0$$

So, by second derivative test,

$$f$$ has local maxima at $$x=\dfrac{\pi}{6}$$

$$f$$ has local minima at $$x=\dfrac{5\pi}{6}$$

Local maxima value $$=f\left(\dfrac{\pi}{6}\right)=\dfrac{3\sqrt3}{4}$$

Local minimum value $$=f\left(\dfrac{5\pi}{6}\right)=\dfrac{-3\sqrt3}{4}$$

Find the local maximum and local minimum values of $$f(x)=sin\,x(1+cos\,x)$$ by second derivative test $$(0<x<2\,\pi)$$.

A Local maximum value $$=f(0) =0$$ , Local minimum value $$=f\left(\dfrac{3\pi}{2}\right)=-1$$

B Local maximum value $$=f\left(\dfrac{\pi}{3}\right) =\dfrac{3\sqrt3}{4}$$ , Local minimum value $$=f\left(\dfrac{5\pi}{3}\right)=\dfrac{-3\sqrt3}{4}$$

C Local maximum value $$=f\left(\pi\right) =0$$ , Local minimum value $$=f\left(\dfrac{5\pi}{3}\right)=\dfrac{-3\sqrt3}{4}$$

D Local maximum value $$=f\left(\dfrac{\pi}{3}\right) =\dfrac{3\sqrt3}{4}$$ , Local minimum value $$=f\left(0\right)=0$$

×

$$f'(x) = cos \,x (1+cos\,x) + sin \,x -sin\,x$$

$$= cos\,x +cos^2\,x -sin^2\,x$$

$$= cos\,x+cos^2\,x -1 +cos^2\,x$$

$$= 2\,cos^2\,x +cos\,x-1$$

$$f'(x)=0\,\Rightarrow 2\,cos^2\,x +cos\,x-1=0$$

$$\Rightarrow (cos\,x+1)(2\,cos\,x-1)=0$$

$$\Rightarrow cos\,x=-1,\dfrac{1}{2}$$

$$\Rightarrow x= \pi,\dfrac{\pi}{3},\dfrac{5\pi}{3}\to$$ critical point.

Now

$$f'' (x)= \dfrac{d}{dx} (2\,cos^2\,x+cos\,x-1)$$

$$= -4 \,cosx\,sin\,x-sin\,x$$

$$f'' (0) =0$$

$$f''\left(\dfrac{\pi}{3}\right)= -4× \dfrac{1}{2}× \dfrac{\sqrt3}{2} - \dfrac{\sqrt3}{2} <0$$

$$f''\left(\dfrac{5\pi}{3}\right)= -4× \dfrac{1}{2}× \dfrac{-\sqrt3}{2} + \dfrac{\sqrt3}{2} >0$$

By second derivative  test

$$f$$ has local maxima at $$x = \dfrac{\pi}{3}$$

$$f$$ has local minima at $$x = \dfrac{5\pi}{3}$$

Local maximum value $$=f\left(\dfrac{\pi}{3}\right) =\dfrac{3\sqrt3}{4}$$ ,

Local minimum value $$=f\left(\dfrac{5\pi}{3}\right)=\dfrac{-3\sqrt3}{4}$$

Find the local maximum and local minimum values of $$f(x)=sin\,x(1+cos\,x)$$ by second derivative test $$(0<x<2\,\pi)$$.

A

Local maximum value $$=f(0) =0$$ , Local minimum value $$=f\left(\dfrac{3\pi}{2}\right)=-1$$

.

B

Local maximum value $$=f\left(\dfrac{\pi}{3}\right) =\dfrac{3\sqrt3}{4}$$ , Local minimum value $$=f\left(\dfrac{5\pi}{3}\right)=\dfrac{-3\sqrt3}{4}$$

C

Local maximum value $$=f\left(\pi\right) =0$$ , Local minimum value $$=f\left(\dfrac{5\pi}{3}\right)=\dfrac{-3\sqrt3}{4}$$

D

Local maximum value $$=f\left(\dfrac{\pi}{3}\right) =\dfrac{3\sqrt3}{4}$$ , Local minimum value $$=f\left(0\right)=0$$

Option B is Correct

Point of Inflection

• A points  P on a curve  $$y=f(x)$$ is called an inflection point  if $$f$$ is continuous at that point and the curve changes from concave upward to concave downward or from concave downwards  to concave upward at P .

• Here points B and C are points of inflection .
• In general an inflection point is a point where a curve changes its direction of concavity.

Identify points of inflection on the following curve.

A $$A,C$$

B $$B$$

C $$C$$

D $$D,E$$

×

An inflection point is a point where a curve changes its direction of concavity.

Hence , Points $$D,E$$ are points of inflection .

Identify points of inflection on the following curve.

A

$$A,C$$

.

B

$$B$$

C

$$C$$

D

$$D,E$$

Option D is Correct

Inflection Points by using Second Derivative Test

• A point  on curve  $$y= f(x)$$ is called  point of inflection if $$'f'$$ is continuous there and  curve changes  concavity at this point  i.e  either  it changes  from concave upwards to concave downward or concave downward to concave upwards.
• 'C' is a point of inflection in these graphs.

• If curve has a tangent at point of inflection, then  curve crosses its tangent there.
• There is a point of inflection at any point where second derivative changes sign (either from positive to negative or negative to positive ).
• For inflection points $$\to \dfrac{d^2\,y}{dx} = \underbrace {f''(x) =0 }_{\text{necessary action }}$$ (if it exists)

For example:

We will find inflection points for the curve $$y=2x^4-3x^3+4x+6$$

For points of inflection $$\to$$  $$f''(x)=\dfrac{d^2y}{dx^2}=0$$ (necessary condition)

Step 1: Find $$f'(x)$$

$$f'(x)=\dfrac{d}{dx}(2x^4-3x^3+4x+6)$$

$$f'(x)=8x^3-9x^2+4$$

Step 2: Find $$f''(x)$$

$$f''(x)=\dfrac{d}{dx}(f'(x))$$

$$f''(x)=\dfrac{d}{dx}(8x^3-9x^2+4)$$

$$f''(x)=24x^2-18x$$

Step 3: Put $$f''(x)=0$$ and solve for $$x.$$

$$f''(x)=0$$

$$24x^2-18x=0$$

$$6x(4x-3)=0$$

$$x=0,\;\dfrac{3}{4}$$

Step 4: Make a table for the sign

Interval $$4x-3$$ $$x$$ $$f''$$
$$x>\dfrac{3}{4}$$ + + +
$$0<x<\dfrac{3}{4}$$ +
$$x<0$$ +

$$f''(x)$$ changes sign from positive to negative at $$x=\dfrac{3}{4}$$.

So, $$x=\dfrac{3}{4}$$ is a point of inflection.

$$f''(x)$$ changes sign from negative to positive at $$x=0$$.

So, $$x=0$$ is also a point of inflection.

Find the $$x$$ coordinate points  of inflection  for the  curve  $$y= 3\,x^5-5\,x^4+3\,x-2$$ .

A $$x = 0$$

B $$x = 1$$

C $$x = 2$$

D $$x = -1$$

×

For points of inflection $$\to f''(x) = \dfrac{d^2y}{d\,x^2}=0$$ (necessary condition )

$$f'(x)=\dfrac{d\,y}{d\,x} = 15\,x^4 - 20\,x^3+3$$

$$f''(x)=\dfrac{d^2\,y}{d\,x^2} = 60\,x^3 - 60\,x^2= 60\,x^2(x-1)$$

$$=f''(x)=0$$

$$\Rightarrow 60\,x^2(x-1)=0 \Rightarrow x=0,1$$

 Interval $$x -1$$ $$x^2$$ $$f''$$ $$x >1$$ $$+$$ $$+$$ $$+$$ $$0 \(f''(x)$$ changes  sign  from negative  to positive  as curve crosses  $$x =1\Rightarrow x=1$$  is point of  inflection.

$$f''(x)$$ doesn't  change  sign  as it crosses  $$x =0 \Rightarrow x= 0$$  is not  a point of inflection .

Find the $$x$$ coordinate points  of inflection  for the  curve  $$y= 3\,x^5-5\,x^4+3\,x-2$$ .

A

$$x = 0$$

.

B

$$x = 1$$

C

$$x = 2$$

D

$$x = -1$$

Option B is Correct