• 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.

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 \(f'(x)=0\)

\(f'(x)=0\)

\(6x^2-6=0\)

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

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

\(x=1,\;-1\)

Step 3: Check the sign in the intervals.

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 \(f\)

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, \(\boxed{a=-8}\)

• 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

From the table, we have

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

and concave downwards on \((1,\;2)\)

• 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 \(=f\left(\dfrac{5\pi}{6}\right)=\dfrac{-3\sqrt3}{4}\)

• 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.  

• 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

\(f''(x)\) 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.