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Wize AP Calculus (BC) Textbook > Applications of Differentiation: Analytical

Maximum and Minimum on Closed Intervals

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Maximum and Minimum on Closed Intervals

A continuous function on a closed interval will always have a largest and smallest value. The values are called Absolute Extrema or Global Extrema.

Absolute Extrema

A function f(x)f(x) has an absolute maximum at a point xx in its domain if f(x)≥f(y) f(x)\ge f(y) for every yy in the domain.



A function f(x)f(x) has an absolute minimum at a point xx in its domain if f(x)≤f(y)f(x) \le f(y)for every yy in the domain.



Note: This definition can also be applied to a function on an interval. The absolute extrema are the biggest and smallest value achieved by the function on that interval.

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Maximum and Minimum on Closed Intervals

For a continuous function f(x)f(x) on a closed interval [a,b],[a, b], there exists an absolute minimum and an absolute maximum.

Watch Out!
This is not necessarily true on an open interval!

Procedure to find the Absolute Extrema of a function on a Closed Interval [a,b]

  1. Find all critical points and singular points of f(x)f(x) in the open interval (a,b)(a, b)
  2. Compute f(x)f(x) at the critical points and singular points and at the endpoints x=ax = a and x=bx = b
  3. The largest and smallest values found are the absolute maximum and minimum respectively
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Example: Finding Absolute Extrema

Find the absolute extrema (max/min) of f(x)=e−x2+1f\left(x\right)=e^{-x^2}+1 on the interval [−1, ∞)\left[-1,\ \infty\right).

1. Find all critical points of the function
The first derivative is f′(x)=e−x2(−2x)f'\left(x\right)=e^{-x^2}\left(-2x\right).
f′(x)=0   →   e−x2(−2x)=0   →   x=0f'\left(x\right)=0\ \ \ \to\ \ \ e^{-x^2}\left(-2x\right)=0\ \ \ \to\ \ \ x=0

2. Compute the yy value of the critical points
*If an interval is given, compute the yy values at these end points

f(0)=e−02+1=2f\left(0\right)=e^{-0^2}+1=2

f(−1)=e−(−1)2+1=e−1+1f\left(-1\right)=e^{-\left(-1\right)^2}+1=e^{-1}+1

"Sub" in the right end-point:

f(∞)→e−(∞)2+1=e−∞+1=0+1=1f\left(\infty\right)\to e^{-\left(\infty\right)^2}+1=e^{-\infty}+1=0+1=1


3. The largest and smallest values that we find are the absolute max and min, respectively
Observe that 1<e−1+1<21<e^{-1}+1<2.

Therefore, the absolute max point is (0, 2)\left(0,\ 2\right) and there is no absolute min point, but instead, as x→∞x\to\infty, there is a horizontal asymptote at y=1y=1 (all values are above this line)
Find the absolute minimum of f(x)=arcsin(x2)f(x)=\text{arcsin}\left(\frac{x}{2}\right) on the interval [1,2][1,2].
Find the absolute maximum and minimum of value of f(x)=2xx2+1\displaystyle f(x)=\frac{2x}{x^2+1} on the interval [0,2][0,2]