Practice first and second derivative test for local extrema, find local maxima and local minima of a function by second derivative test & concavity test.
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.
Let C be a critical number of a continuous function \(f\)
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 \(f\)
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 = \(f(-1) =-1\)
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 = \(f(0) =1\)
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
Note : That there can be more than one parameter in same cases.
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, \(\boxed{a=-8}\)
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 \(f(x)= \,x^3 - 6\,x^2 +9\,x+15\)
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.
(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 \((1,\;2)\)
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 \((0,2)\)
If \(f''\) is continuous near 'C', and \(f'(C) =0 \,\,\,{\text &} \,\,f'' (C) <0\Rightarrow f\) has a local maxima 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 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 \(=f\left(0\right)=0\)
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.
A \(x = 0\)
B \(x = 1\)
C \(x = 2\)
D \(x = -1\)