notes/education/math/MATH1210 (calc 1)/Max and Min.md

z# Maximum/Minimum A function f has an absolute maximum at c if f(c) >= f(x). We call f(c) the maximum value of f. The absolute maximum is the largest possible output value for a function.

A function f has an absolute minimum at c if f(c) <= f(x). f(c) is the absolute minimum value of f. The absolute minimum is the smallest possible output value for a function.

• Where the derivative of a function is zero, there is either a peak or a trough.

Critical Numbers

A number is considered critical if the output of a function exists and \dfrac{d}{dx} is zero or undefined.

Local Max/Min

A local max/min is a peak or trough at any point along the graph.

Extreme Value Theorem

If f is a continuous function in a closed interval [a, b], then f achieves both an absolute maximum and an absolute minimum in [a, b]. Furthermore, the absolute extrema occur at a or at b or at a critical number between a and b.

Examples

Find the absolute maximum and absolute minimum of the function f(x) = x^2 -3x + 2 on the closed interval [0, 2]:

1. x=0 and x=2 are both critical numbers because they are endpoints. Endpoints are always critical numbers because \dfrac{d}{dx} is undefined.
2. \dfrac{d}{dx} x^2 -3x + 2 = 2x -3
3. Setting the derivative to zero, 0 = 2x - 3
4. Solving for x, we get x = \dfrac{3}{2}. Three halves is a critical number because f'(\dfrac{3}{2}) is 0.
5. Now check the outputs for all critical numbers (f(x) at x = \{0, 2, \dfrac{3}{2}\})
6. f(0) = 0^2 -3(0) + 2 = 2
7. f(2) = 2^2 - 3(2) + 2) = 0
8. f(\dfrac{3}{2}) = (\dfrac{3}{2})^2 - 3(\dfrac{3}{2}) + 2 = -\dfrac{1}{4}
9. The minimum is the lowest of the three, so it's -\dfrac{1}{4} and it occurs at x = \dfrac{3}{2}
10. The maximum is the highest of the three, so it's 2 at x = 0.

Find the absolute maximum and absolute minimum of the function h(x) = x + 2cos(x) on the closed interval [0, \pi].

1. x = 0 and x = \pi are both critical numbers because they are endpoints. Endpoints are critical because \dfrac{d}{dx} is undefined.
2. \dfrac{d}{dx} x + 2\cos(x) = 1 - 2sin(x)
3. Setting that to zero, we get 0 = 1 - 2\sin(x)
4. \sin(x) = \dfrac{1}{2}
5. In the interval [0, \pi], \sin(x) has a value of \dfrac{1}{2} in two places: x = \dfrac{\pi}{6} and x = \dfrac{5\pi}{6}. These are both critical numbers because they are points where \dfrac{d}{dx} is zero.
6. Now we plug these values into the original function:
7. h(0) = 0 + 2\cos 0 = 2
8. h(\pi) = \pi + 2\cos(\pi) = \pi - 2 \approx 1.14159
9. h(\dfrac{\pi}{6}) = \dfrac{\pi}{6} + 2\cos(\dfrac{\pi}{6}) = \dfrac{\pi}{6} + 2(\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{6} + \sqrt{3} \approx 2.2556
10. h(\dfrac{5\pi}{6}) = \dfrac{5\pi}{6} + 2\cos(\dfrac{5\pi}{6}) = \dfrac{5\pi}{6} - \sqrt{3} \approx 0.88594