Wize High School Grade 12 Calculus Textbook > Curve Sketching

Critical Points & Local Max/Min

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Critical Point
Given a function f(x)f\left(x\right)f(x), (c, f(c))\left(c,\ f\left(c\right)\right)(c, f(c)) is a critical point if f′(c)=0f'\left(c\right)=0f′(c)=0 or f′(c)f'\left(c\right)f′(c) is undefined.
The value(s) of ccc are called the critical numbers.

The function may turn around at a critical point, so sometimes (c, f(c))\left(c,\ f\left(c\right)\right)(c, f(c)) is called the turning point.

Local Maximum & Minimum
f(a)f\left(a\right)f(a) is a local maximum if f(a)f\left(a\right)f(a) is the largest value on a small interval around x=ax=ax=a
f(b)f\left(b\right)f(b) is a local minimum if f(b)f\left(b\right)f(b) is the smallest value on a small interval around x=bx=bx=b

Watch Out!
If f(a)f\left(a\right)f(a) is a local maximum, it doesn't mean that it is the maximum value of the function on a closed interval

If f(b)f\left(b\right)f(b) is a local minimum, it doesn't mean that it is the minimum value of the function on a closed interval


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Example
Identify all of the critical points, local max/min and global max/min points of the following function on the interval −1≤x≤5-1\le x\le5−1≤x≤5.

Critical points: (1, 1)\left(1,\ 1\right)(1, 1) and (3, 0)\left(3,\ 0\right)(3, 0)
Local maximum: (1, 1)\left(1,\ 1\right)(1, 1)
Local minimum: (3, 0)\left(3,\ 0\right)(3, 0)
Absolute maximum: (5, 5)\left(5,\ 5\right)(5, 5)
Absolute minimum: (1, −4)\left(1,\ -4\right)(1, −4)

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How to find the local max/min? (First derivative test)
1. Find the critical numbers → set f′(x)=0f'\left(x\right)=0f′(x)=0
2. Put the critical numbers on a number line → check if f′f'f′ is positive or negative on the left and right of these critical numbers
3. Make a conclusion based on this table
To the left of cTo the right of cConclusionf′<0f′>0c is a local minimumf′>0f′<0c is a local maximumotherwiseIf f′ doesn’t change changec is neither a local max or min\begin{array}{|c|c|c|}
\hline
\text{To the left of }c&\text{To the right of }c&\text{Conclusion}\\

\hline
f'<0&f'>0&c\text{ is a local minimum}\\

\hline
f'>0&f'<0&c\text{ is a local maximum}\\
\hline
\text{otherwise}&\text{If }f'\text{ doesn't change change}&c\text{ is neither a local max or min}\\
\hline
\end{array}To the left of cf′<0f′>0otherwise​To the right of cf′>0f′<0If f′ doesn’t change change​Conclusionc is a local minimumc is a local maximumc is neither a local max or min​​

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Example: Finding Local Extrema
Find and classify all local max and min of the graph of f(x)=x3−9x2+15xf\left(x\right)=x^3-9x^2+15xf(x)=x3−9x2+15x.

1. Find the critical numbers:
f′(x)=3x2−18x+15f'\left(x\right)=3x^2-18x+15f′(x)=3x2−18x+15
f′(x)=3(x2−6x+5)f'\left(x\right)=3\left(x^2-6x+5\right)f′(x)=3(x2−6x+5)
f′(x)=3(x−5)(x−1)f'\left(x\right)=3\left(x-5\right)\left(x-1\right)f′(x)=3(x−5)(x−1)
Set to 0:
0=3(x−5)(x−1)0=3\left(x-5\right)\left(x-1\right)0=3(x−5)(x−1)
x=1 or 5x=1\ or\ 5x=1 or 5

2. Put the critical numbers on a number line:

3. Make a conclusion:
x=1x=1x=1: increasing on the left, decreasing on the right → it corresponds to a local max
x=5x=5x=5: decreasing on the left, increasing on the right → it corresponds to a local min
Therefore, (1, 7)\left(1,\ 7\right)(1, 7) is a local max and (5, −25)\left(5,\ -25\right)(5, −25) is a local min.

Practice: Finding Local Extrema
Find and classify all local max and min of the graph of y=x2−x+1y=\sqrt{x^2-x+1}y=x2−x+1​.

Local min: None
Local max: (12, 32)\left(\frac{1}{2},\ \frac{\sqrt{3}}{2}\right)(21​, 23​​)

Local min: (12, 32)\left(\frac{1}{2},\ \frac{\sqrt{3}}{2}\right)(21​, 23​​)
Local max: None

Local min: (12, 32)\left(\frac{1}{2},\ \frac{\sqrt{3}}{2}\right)(21​, 23​​)
Local max: (1, 1)\left(1,\ 1\right)(1, 1)

Local min: (1, 1)\left(1,\ 1\right)(1, 1)
Local max: (12, 32)\left(\frac{1}{2},\ \frac{\sqrt{3}}{2}\right)(21​, 23​​)

None of the above

I don't know

Practice: Finding Local Extrema
Find and classify all local max and min of the graph of y=6x4+4x3y=6x^4+4x^3y=6x4+4x3.

Local min: None
Local max: (−12, −18)\left(-\frac{1}{2},\ -\frac{1}{8}\right)(−21​, −81​)

Local min: (−12, −18)\left(-\frac{1}{2},\ -\frac{1}{8}\right)(−21​, −81​)
Local max: None

Local min: (−12, −18)\left(-\frac{1}{2},\ -\frac{1}{8}\right)(−21​, −81​)
Local max: (0, 0)\left(0,\ 0\right)(0, 0)

Local min: (0, 0)\left(0,\ 0\right)(0, 0)
Local max: (−12, −18)\left(-\frac{1}{2},\ -\frac{1}{8}\right)(−21​, −81​)

None of the above

I don't know

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Example: Local Extrema & Graphs
Sketch a graph of a function that is differentiable on the interval −1≤x≤5-1\le x\le5−1≤x≤5 and meets the following criteria:
f′>0f'>0f′>0 on the interval 0<x<30<x<30<x<3
f′<0f'<0f′<0 on the interval −1<x<0-1<x<0−1<x<0 and 3<x<53<x<53<x<5
The graph has a local extrema at (0, 2)\left(0,\ 2\right)(0, 2) and (3, 10)\left(3,\ 10\right)(3, 10)
On the interval −1≤x≤5-1\le x\le5−1≤x≤5, the absolute maximum is 15 and the absolute minimum is 2

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Example: Local Extrema
Find the values of a, b, ca,\ b,\ ca, b, c and ddd such that the function f(x)=ax3+bx2+cx+df\left(x\right)=ax^3+bx^2+cx+df(x)=ax3+bx2+cx+d has a local maximum at (−2, 21)\left(-2,\ 21\right)(−2, 21) and a local minimum at (1, −6)\left(1,\ -6\right)(1, −6).

Sub in the points:
We know that the function contains the point (−2,21)(-2, 21)(−2,21):
21=a(−2)3+b(−2)2+c(−2)+d21=a\left(-2\right)^3+b\left(-2\right)^2+c\left(-2\right)+d21=a(−2)3+b(−2)2+c(−2)+d
21=−8a+4b−2c+d\boxed{21=-8a+4b-2c+d}21=−8a+4b−2c+d​ - equation 1

We know that the function contains the point (1, −6)\left(1,\ -6\right)(1, −6):
−6=a(1)3+b(1)2+c(1)+d-6=a\left(1\right)^3+b\left(1\right)^2+c\left(1\right)+d−6=a(1)3+b(1)2+c(1)+d
−6=a+b+c+d\boxed{-6=a+b+c+d}−6=a+b+c+d​ - equation 2

Find the critical number(s):
f′(x)=3ax2+2bx+cf'\left(x\right)=3ax^2+2bx+cf′(x)=3ax2+2bx+c

The local max is at x=−2x=-2x=−2 and the local min is at x=1x=1x=1
0=3a(−2)2+2b(−2)+c →0=12a−4b+c0=3a\left(-2\right)^2+2b\left(-2\right)+c\ \to \boxed{0=12a-4b+c}0=3a(−2)2+2b(−2)+c →0=12a−4b+c​ - equation 3
0=3a(1)2+2b(1)+c → 0=3a+2b+c0=3a(1)^2+2b(1)+c\ \to\ \boxed{0=3a+2b+c}0=3a(1)2+2b(1)+c → 0=3a+2b+c​ - equation 4
Combining equations 1-4:
From equation 4: d=21+8a−4b+2cd=21+8a-4b+2cd=21+8a−4b+2c
Sub this into equations 2:
−6=a+b+c+(21+8a−4b+2c)-6=a+b+c+\left(21+8a-4b+2c\right)−6=a+b+c+(21+8a−4b+2c)
−27=9a−3b+3c-27=9a-3b+3c−27=9a−3b+3c
−9=3a−b+c-9=3a-b+c−9=3a−b+c
c=−9−3a+b\boxed{c=-9-3a+b}c=−9−3a+b​
Sub this into equations 3 and 4:
0=12a−4b+(−9−3a+b) → 9=9a−3b → 3=3a−b0=12a-4b+(-9-3a+b)\ \to\ 9=9a-3b\ \to\ \boxed{3=3a-b}0=12a−4b+(−9−3a+b) → 9=9a−3b → 3=3a−b​
0=3a+2b+(−9−3a+b) → 9=3b → b=30=3a+2b+(-9-3a+b)\ \to\ 9=3b\ \to\ \boxed{b=3}0=3a+2b+(−9−3a+b) → 9=3b → b=3​
So, b=3\boxed{b=3}b=3​ and a=2\boxed{a=2}a=2​.
Solve for c and d: 
c=−9−3(2)+(3) → c=−12c=-9-3\left(2\right)+\left(3\right)\ \to\ \boxed{c=-12}c=−9−3(2)+(3) → c=−12​
d=21+8(2)−4(3)+2(−12) → d=1d=21+8\left(2\right)-4\left(3\right)+2\left(-12\right)\ \to\ \boxed{d=1}d=21+8(2)−4(3)+2(−12) → d=1​

Therefore, a=2, b=3, c=−12, d=1a=2,\ b=3,\ c=-12,\ d=1a=2, b=3, c=−12, d=1

Find the value(s) of ccc such that f(x)=cx3−3(c+1)xf\left(x\right)=cx^3-3\left(c+1\right)xf(x)=cx3−3(c+1)x has