To find the absolute maximum and minimum of a function using derivatives, you must first identify all critical points—the points where the derivative is zero or undefined. The process depends on whether the function is defined on a closed interval or an open interval.
Finding absolute extrema on a closed interval [a,b]open bracket a comma b close bracket
[𝑎,𝑏]
The Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval will have an absolute maximum and minimum. These extreme values must occur at either a critical point within the interval or at one of the endpoints.
The Closed Interval Method
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Check continuity: Confirm that your function f(x)f of x
𝑓(𝑥)
is continuous on the closed interval [a,b]open bracket a comma b close bracket
[𝑎,𝑏]
. Polynomials, exponential, sine, and cosine functions are continuous everywhere.
-
Find critical points: Calculate the first derivative, f′(x)f prime of x
𝑓′(𝑥)
, and find the values of xx
𝑥
for which f′(x)=0f prime of x equals 0
𝑓′(𝑥)=0
or f′(x)f prime of x
𝑓′(𝑥)
is undefined.
-
Filter critical points: Discard any critical points that fall outside the given interval (a,b)open paren a comma b close paren
(𝑎,𝑏)
.
-
Evaluate at all candidate points: Calculate the value of the original function, f(x)f of x
𝑓(𝑥)
, for every critical point found in step 2 and for both endpoints, aa
𝑎
and bb
𝑏
.
-
Compare values: The largest value from step 4 is the absolute maximum, and the smallest is the absolute minimum.
Example
Find the absolute maximum and minimum of the function f(x)=x3−3x2+1f of x equals x cubed minus 3 x squared plus 1
𝑓(𝑥)=𝑥3−3𝑥2+1
on the interval [-2,4]open bracket negative 2 comma 4 close bracket
[−2,4]
.
-
Continuity: f(x)f of x
𝑓(𝑥)
is a polynomial, so it is continuous everywhere.
-
Critical points:
-
f′(x)=3x2−6xf prime of x equals 3 x squared minus 6 x
𝑓′(𝑥)=3𝑥2−6𝑥
-
Set f′(x)=0⟹3x(x−2)=0f prime of x equals 0 ⟹ 3 x open paren x minus 2 close paren equals 0
𝑓′(𝑥)=0⟹3𝑥(𝑥−2)=0
-
The critical points are x=0x equals 0
𝑥=0
and x=2x equals 2
𝑥=2
.
-
-
Filter: Both x=0x equals 0
𝑥=0
and x=2x equals 2
𝑥=2
are inside the interval [-2,4]open bracket negative 2 comma 4 close bracket
[−2,4]
.
-
Evaluate:
-
f(-2)=(-2)3−3(-2)2+1=-8−12+1=-19f of negative 2 equals open paren negative 2 close paren cubed minus 3 open paren negative 2 close paren squared plus 1 equals negative 8 minus 12 plus 1 equals negative 19
𝑓(−2)=(−2)3−3(−2)2+1=−8−12+1=−19
-
f(0)=(0)3−3(0)2+1=1f of 0 equals open paren 0 close paren cubed minus 3 open paren 0 close paren squared plus 1 equals 1
𝑓(0)=(0)3−3(0)2+1=1
-
f(2)=(2)3−3(2)2+1=8−12+1=-3f of 2 equals open paren 2 close paren cubed minus 3 open paren 2 close paren squared plus 1 equals 8 minus 12 plus 1 equals negative 3
𝑓(2)=(2)3−3(2)2+1=8−12+1=−3
-
f(4)=(4)3−3(4)2+1=64−48+1=17f of 4 equals open paren 4 close paren cubed minus 3 open paren 4 close paren squared plus 1 equals 64 minus 48 plus 1 equals 17
𝑓(4)=(4)3−3(4)2+1=64−48+1=17
-
-
Compare: The largest value is 17 (at x=4x equals 4
𝑥=4
) and the smallest is -19 (at x=-2x equals negative 2
𝑥=−2
).
- Absolute maximum: 17
- Absolute minimum: -19
Finding absolute extrema on an open interval (a,b)open paren a comma b close paren
(𝑎,𝑏)
On an open interval, there is no guarantee that an absolute maximum or minimum exists. You must analyze the behavior of the function at its critical points and at the endpoints using limits.
Method for open intervals (a,b)open paren a comma b close paren
(𝑎,𝑏)
-
Find critical points: Find the critical points in the open interval (a,b)open paren a comma b close paren
(𝑎,𝑏)
where f′(x)=0f prime of x equals 0
𝑓′(𝑥)=0
or f′(x)f prime of x
𝑓′(𝑥)
is undefined.
-
Analyze end behavior: Find the limits of the function as xx
𝑥
approaches the endpoints.
-
limx→a+f(x)limit over x right arrow a raised to the positive power of f of x
lim𝑥→𝑎+𝑓(𝑥)
-
limx→b−f(x)limit over x right arrow b raised to the negative power of f of x
lim𝑥→𝑏−𝑓(𝑥)
-
-
Compare values:
-
Evaluate f(x)f of x
𝑓(𝑥)
at each critical point.
-
Compare the function values at the critical points with the limits at the endpoints.
-
If the value of a critical point is greater than both endpoint limits, it is the absolute maximum. If it is less than both, it is the absolute minimum.
-
If a limit approaches ∞infinity
∞
, there is no absolute maximum. If a limit approaches −∞negative infinity
−∞
, there is no absolute minimum.
-
Example
Find the absolute extrema of f(x)=x2x2+1f of x equals the fraction with numerator x squared and denominator x squared plus 1 end-fraction
𝑓(𝑥)=𝑥2𝑥2+1
on the interval (−∞,∞)open paren negative infinity comma infinity close paren
(−∞,∞)
.
- Critical points:
-
Using the quotient rule, f′(x)=(x2+1)(2x)−(x2)(2x)(x2+1)2=2x(x2+1)2f prime of x equals the fraction with numerator open paren x squared plus 1 close paren open paren 2 x close paren minus open paren x squared close paren open paren 2 x close paren and denominator open paren x squared plus 1 close paren squared end-fraction equals the fraction with numerator 2 x and denominator open paren x squared plus 1 close paren squared end-fraction
𝑓′(𝑥)=(𝑥2+1)(2𝑥)−(𝑥2)(2𝑥)(𝑥2+1)2=2𝑥(𝑥2+1)2
.
-
Set f′(x)=0⟹2x=0⟹x=0f prime of x equals 0 ⟹ 2 x equals 0 ⟹ x equals 0
𝑓′(𝑥)=0⟹2𝑥=0⟹𝑥=0
.
-
The only critical point is x=0x equals 0
𝑥=0
.
-
- End behavior:
-
limx→−∞x2x2+1=limx→−∞11+1/x2=1limit over x right arrow negative infinity of the fraction with numerator x squared and denominator x squared plus 1 end-fraction equals limit over x right arrow negative infinity of the fraction with numerator 1 and denominator 1 plus 1 / x squared end-fraction equals 1
lim𝑥→−∞𝑥2𝑥2+1=lim𝑥→−∞11+1/𝑥2=1
.
-
limx→∞x2x2+1=limx→∞11+1/x2=1limit over x right arrow infinity of the fraction with numerator x squared and denominator x squared plus 1 end-fraction equals limit over x right arrow infinity of the fraction with numerator 1 and denominator 1 plus 1 / x squared end-fraction equals 1
lim𝑥→∞𝑥2𝑥2+1=lim𝑥→∞11+1/𝑥2=1
.
-
- Compare:
-
The function value at the critical point is f(0)=0202+1=0f of 0 equals the fraction with numerator 0 squared and denominator 0 squared plus 1 end-fraction equals 0
𝑓(0)=0202+1=0
.
-
The value of f(0)=0f of 0 equals 0
𝑓(0)=0
is less than the limit value of 1.
-
Since f(x)f of x
𝑓(𝑥)
is always non-negative (because x2≥0x squared is greater than or equal to 0
𝑥2≥0
), the value at the critical point is the lowest value on the interval.
-
The function never actually reaches the value of 1, so there is no absolute maximum.
-
Absolute minimum: 0 (at x=0x equals 0
𝑥=0
)
-
Absolute maximum: None
-