Lecture: Section 4.1 Maximum and Minimum Values

Solution:

• Absolute Maximum: The highest point on the entire graph occurs at the endpoint $x=-4$, where the value is $f(-4) = 97$.
• Absolute Minimum: The lowest point on the entire graph occurs at the endpoint $x=4$, where the value is $f(4) = -31$.
• Local Maximums: There is a "peak" at $x \approx 1.33$ (or $4/3$), with value $f(4/3) \approx 2.185$. The endpoint $x=-4$ is also considered a local maximum.
• Local Minimums: There is a "valley" at $x=0$, with value $f(0) = 1$. The endpoint $x=4$ is also a local minimum.

Solution:

The function has an absolute minimum of $f(-2)=-5$. However, it has no absolute maximum.

As $x$ approaches 1, the value of $f(x)$ gets closer and closer to 4, but it never reaches 4 because of the hole in the graph at that point. The defined point $f(1)=1$ is not the maximum value.

This does not contradict the EVT. The theorem requires the function to be continuous on the entire closed interval. This function has a removable discontinuity at $x=1$. Since the conditions of the theorem are not met, it is not guaranteed to have an absolute maximum and minimum, and in this case, it doesn't have a maximum.

Solution:

Step 1: Find critical numbers.
$g'(t) = -2\sin(t) + 2\cos(2t)$. Using the identity $\cos(2t) = 1-2\sin^2(t)$, we get:
$g'(t) = -2\sin(t) + 2(1-2\sin^2(t)) = -4\sin^2(t) - 2\sin(t) + 2$.
Set $g'(t)=0$ and let $u=\sin(t)$: $-4u^2 - 2u + 2 = 0 \implies 2u^2 + u - 1 = 0$.
Factoring gives $(2u-1)(u+1) = 0$, so $u=1/2$ or $u=-1$.
This means $\sin(t)=1/2$ or $\sin(t)=-1$.
In the interval $[0, \pi/2]$, the only solution is $\sin(t)=1/2 \implies t=\pi/6$.

Step 2 & 3: Evaluate $g(t)$ at critical numbers and endpoints.

• (Critical Number) $g(\pi/6) = 2\cos(\pi/6) + \sin(\pi/3) = 2(\frac{\sqrt{3}}{2}) + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \approx 2.598$.
• (Endpoint) $g(0) = 2\cos(0) + \sin(0) = 2(1) + 0 = 2$.
• (Endpoint) $g(\pi/2) = 2\cos(\pi/2) + \sin(\pi) = 2(0) + 0 = 0$.

Step 4: Compare. The values are $\{ \frac{3\sqrt{3}}{2}, 2, 0 \}$. The largest is $\frac{3\sqrt{3}}{2}$ and the smallest is 0.

Conclusion: The absolute maximum is $\frac{3\sqrt{3}}{2}$ (at $t=\pi/6$) and the absolute minimum is 0 (at $t=\pi/2$).

Solution:

Step 1: Determine the domain. The function is defined when $8-x^2 \ge 0$, which means $x^2 \le 8$. So, the domain is a closed interval: $[-\sqrt{8}, \sqrt{8}]$, or $[-2\sqrt{2}, 2\sqrt{2}]$. Since the function is continuous on a closed interval, the EVT guarantees absolute extrema exist.

Step 2: Find the derivative. Using the product rule:
$f'(x) = 1 \cdot \sqrt{8-x^2} + x \cdot \frac{1}{2\sqrt{8-x^2}}(-2x)$
$f'(x) = \sqrt{8-x^2} - \frac{x^2}{\sqrt{8-x^2}}$
Combine into a single fraction: $f'(x) = \frac{(8-x^2) - x^2}{\sqrt{8-x^2}} = \frac{8-2x^2}{\sqrt{8-x^2}}$.

Step 3: Find critical numbers.
Set the numerator to zero: $8-2x^2 = 0 \implies x^2=4 \implies x = \pm 2$. Both are in the domain.
Find where $f'(x)$ DNE by setting the denominator to zero: $\sqrt{8-x^2}=0 \implies x = \pm \sqrt{8} = \pm 2\sqrt{2}$. These are the endpoints of the domain.

Step 4: Apply the Closed Interval Method.
The candidates are the critical numbers ($x=\pm 2$) and the endpoints ($x=\pm 2\sqrt{2}$).

• $f(-2\sqrt{2}) = (-2\sqrt{2})\sqrt{8 - (2\sqrt{2})^2} = 0$
• $f(-2) = (-2)\sqrt{8 - (-2)^2} = -2\sqrt{4} = -4$
• $f(2) = (2)\sqrt{8 - (2)^2} = 2\sqrt{4} = 4$
• $f(2\sqrt{2}) = (2\sqrt{2})\sqrt{8 - (2\sqrt{2})^2} = 0$

Conclusion: The absolute maximum is 4 (at $x=2$) and the absolute minimum is -4 (at $x=-2$).

Solution:

Step 1: Find the derivative. The function is continuous on the closed interval $[0, 4]$.
$f'(x) = 1 - 2 \cdot \frac{1}{1+x^2} = 1 - \frac{2}{1+x^2}$.

Step 2: Find critical numbers.
The derivative exists for all $x$. Set $f'(x)=0$:
$1 = \frac{2}{1+x^2} \implies 1+x^2 = 2 \implies x^2=1$.
This gives $x=\pm 1$. Only $x=1$ is in the interval $[0, 4]$. So, $x=1$ is our only critical number to check inside the interval.

Step 3: Apply the Closed Interval Method.
The candidates are the critical number ($x=1$) and the endpoints ($x=0, x=4$).

• $f(0) = 0 - 2\arctan(0) = 0$
• $f(1) = 1 - 2\arctan(1) = 1 - 2(\frac{\pi}{4}) = 1 - \frac{\pi}{2} \approx -0.571$
• $f(4) = 4 - 2\arctan(4) \approx 4 - 2(1.326) = 1.348$

Step 4: Compare values. The values are $\{0, 1-\pi/2, 4-2\arctan(4)\}$. The smallest is $1-\pi/2$, and the largest is $4-2\arctan(4)$.

Conclusion: The absolute maximum is $4 - 2\arctan(4)$ (which occurs at $x=4$) and the absolute minimum is $1 - \frac{\pi}{2}$ (which occurs at $x=1$).

Solution:

Step 1: Use Fermat's Theorem. For a differentiable function to have a local minimum at $x=-2$, that point must be a critical number. Since the function is differentiable at $x=-2$, we must have $f'(-2)=0$.

Step 2: Find the derivative and solve for $b$.
$f(x) = x^2 + bx^{-1}$
$f'(x) = 2x - bx^{-2} = 2x - \frac{b}{x^2}$.
Now, set $f'(-2)=0$:
$f'(-2) = 2(-2) - \frac{b}{(-2)^2} = -4 - \frac{b}{4} = 0$.
Solving for $b$: $-4 = \frac{b}{4} \implies b = -16$.

Step 3: Find the local minimum value. To find the value, we evaluate the original function $f(x)$ at $x=-2$ using our value for $b$.
$f(x) = x^2 - \frac{16}{x}$
$f(-2) = (-2)^2 - \frac{16}{-2} = 4 - (-8) = 12$.

Conclusion: The constant must be $b=-16$, and the local minimum value is 12.