Application of Derivatives - Maximum and Minimum Values of a Function in a Closed Interval

Find the absolute maximum and minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[1, 5]$.

1. Find the derivative: $f'(x) = 6x^2 - 30x + 36$
2. Set $f'(x) = 0$: $6(x^2 - 5x + 6) = 0 \implies 6(x - 2)(x - 3) = 0$ Thus, critical points are $x = 2$ and $x = 3$. Both lie in the interval $[1, 5]$.
3. Evaluate $f(x)$ at critical points and endpoints:

• $f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24$
• $f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$
• $f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$
• $f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 250 - 375 + 180 + 1 = 56$

1. Compare values:

• Absolute Maximum Value = $56$ at $x = 5$
• Absolute Minimum Value = $24$ at $x = 1$.

To find absolute extrema on a closed interval, we must check the function values at the critical points (where the slope is zero) and the boundaries of the interval. Even if a local maximum exists inside the interval (like at $x=2$), the endpoint value (at $x=5$) might be significantly higher.

Find the absolute maximum and minimum values of $f(x) = \sin x + \cos x$ on $[0, \pi]$.

1. Find $f'(x)$: $f'(x) = \cos x - \sin x$
2. Critical points: $\cos x - \sin x = 0 \implies \tan x = 1 \implies x = \frac{\pi}{4}$ (Since $x \in [0, \pi]$).
3. Evaluate at $x = 0, \frac{\pi}{4}, \pi$:

• $f(0) = \sin 0 + \cos 0 = 0 + 1 = 1$
• $f(\frac{\pi}{4}) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$
• $f(\pi) = \sin \pi + \cos \pi = 0 - 1 = -1$

1. Conclusion:

• Absolute Maximum Value = $\sqrt{2}$ at $x = \frac{\pi}{4}$
• Absolute Minimum Value = $-1$ at $x = \pi$.

In trigonometric functions over a closed interval, the extreme values occur where the components balance out (critical points) or at the constraints of the domain (endpoints).

1. Find $f'(x)$: $f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x - 2)(x + 2)$.
2. Set $f'(x) = 0$: The critical points are $x = 0, x = 2, x = -2$.
3. Check which points lie in $[-3, 1]$: $x = 0$ and $x = -2$ are in the interval. $x = 2$ is outside.
4. Evaluate $f(x)$ at endpoints and valid critical points: $f(-3) = (-3)^4 - 8(-3)^2 + 16 = 81 - 72 + 16 = 25$ $f(-2) = (-2)^4 - 8(-2)^2 + 16 = 16 - 32 + 16 = 0$ $f(0) = (0)^4 - 8(0)^2 + 16 = 16$ $f(1) = (1)^4 - 8(1)^2 + 16 = 1 - 8 + 16 = 9$
5. Compare: Absolute Max is $25$ at $x = -3$; Absolute Min is $0$ at $x = -2$.

We identify critical points within the specified range and compare their functional values against the values at the boundary points $-3$ and $1$.

1. Find $f'(x)$: $f'(x) = 3 - 3x^2 = 3(1 - x^2) = 3(1 - x)(1 + x)$.
2. Set $f'(x) = 0$: $x = 1, x = -1$.
3. Check interval $[0, 2]$: Only $x = 1$ is within the interval.
4. Evaluate values: $f(0) = 3(0) - 0^3 = 0$ $f(1) = 3(1) - 1^3 = 2$ $f(2) = 3(2) - 2^3 = 6 - 8 = -2$
5. Compare: Absolute Maximum is $2$ at $x = 1$; Absolute Minimum is $-2$ at $x = 2$.

Calculus reveals a local peak at $x=1$. By checking the boundaries, we find the function drops below zero at the right endpoint, making it the global minimum.