Derivatives: Learning Objective 8

First, find the derivative, $f'(x)$. It's easiest if you first distribute the $x^{3/5}$ term. Critical numbers are the points in the domain of $f(x)$ where the derivative $f'(x)$ is either zero or undefined.

The critical numbers are $x = 0$ and $x = \frac{3}{2}$.

First, rewrite the function: $$ f(x) = 4x^{3/5} - x^{8/5} $$ Now, find the derivative using the power rule: $$ f'(x) = 4\left(\frac{3}{5}\right)x^{-2/5} - \frac{8}{5}x^{3/5} = \frac{12}{5x^{2/5}} - \frac{8x^{3/5}}{5} $$ To find the critical numbers, we find where $f'(x)=0$ or is undefined. Let's combine the terms into a single fraction: $$ f'(x) = \frac{12 - 8x^{3/5} \cdot x^{2/5}}{5x^{2/5}} = \frac{12 - 8x}{5x^{2/5}} $$ The derivative is zero when the numerator is zero: $$ 12 - 8x = 0 \implies 8x = 12 \implies x = \frac{12}{8} = \frac{3}{2} $$ The derivative is undefined when the denominator is zero: $$ 5x^{2/5} = 0 \implies x = 0 $$ Both $x=0$ and $x=3/2$ are in the domain of the original function $f(x)$, so they are both critical numbers.

Use the Closed Interval Method. First, find the critical numbers by setting the derivative $f'(x)$ to zero. Then, evaluate the function $f(x)$ at these critical numbers and at the endpoints of the interval, $-\frac{1}{2}$ and $4$. The largest value is the absolute maximum and the smallest is the absolute minimum.

The absolute maximum is $17$, which occurs at $x=4$.
The absolute minimum is $-3$, which occurs at $x=2$.

1. Find the derivative: $$ f'(x) = 3x^2 - 6x $$ 2. Find critical numbers: Set $f'(x)=0$. $$ 3x^2 - 6x = 0 \implies 3x(x-2) = 0 $$ The critical numbers are $x=0$ and $x=2$. Both are within the interval $[-\frac{1}{2}, 4]$.

3. Evaluate the function at critical numbers and endpoints:

• $f(-\frac{1}{2}) = (-\frac{1}{2})^3 - 3(-\frac{1}{2})^2 + 1 = -\frac{1}{8} - \frac{3}{4} + 1 = \frac{-1-6+8}{8} = \frac{1}{8}$
• $f(0) = (0)^3 - 3(0)^2 + 1 = 1$
• $f(2) = (2)^3 - 3(2)^2 + 1 = 8 - 12 + 1 = -3$
• $f(4) = (4)^3 - 3(4)^2 + 1 = 64 - 48 + 1 = 17$

4. Compare the values: The largest value is $17$ and the smallest is $-3$.

Apply the Closed Interval Method. Find critical numbers by solving $h'(t)=0$. Compare the function's value at these critical numbers and at the endpoints, $-4$ and $2$.

The absolute maximum is $24$, which occurs at $t=-2$.
The absolute minimum is $-28$, which occurs at $t=-4$.

1. Find the derivative: $$ h'(t) = 6t^2 + 6t - 12 $$ 2. Find critical numbers: Set $h'(t)=0$. $$ 6(t^2 + t - 2) = 0 \implies 6(t+2)(t-1) = 0 $$ The critical numbers are $t=-2$ and $t=1$. Both are within the interval $[-4, 2]$.

3. Evaluate the function at critical numbers and endpoints:

• $h(-4) = 2(-4)^3 + 3(-4)^2 - 12(-4) + 4 = 2(-64) + 3(16) + 48 + 4 = -128 + 48 + 48 + 4 = -28$
• $h(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) + 4 = 2(-8) + 3(4) + 24 + 4 = -16 + 12 + 24 + 4 = 24$
• $h(1) = 2(1)^3 + 3(1)^2 - 12(1) + 4 = 2 + 3 - 12 + 4 = -3$
• $h(2) = 2(2)^3 + 3(2)^2 - 12(2) + 4 = 16 + 12 - 24 + 4 = 8$

4. Compare the values: The largest value is $24$ and the smallest is $-28$.

Use the Product Rule to find the derivative $g'(y)$. Find critical numbers where $g'(y)=0$ or is undefined. Then, apply the Closed Interval Method by testing the critical numbers and endpoints.

The absolute maximum is $3 \cdot 5^{2/3} \approx 8.77$, which occurs at $y=1$.
The absolute minimum is $-\frac{144}{5^{5/3}} \approx -15.06$, which occurs at $y = -12/5$.

1. Find the derivative using the Product Rule: $$ g'(y) = (3)(y+4)^{2/3} + (3y)\left(\frac{2}{3}(y+4)^{-1/3}\right) = 3(y+4)^{2/3} + \frac{2y}{(y+4)^{1/3}} $$ Combine into a single fraction: $$ g'(y) = \frac{3(y+4) + 2y}{(y+4)^{1/3}} = \frac{5y+12}{(y+4)^{1/3}} $$ 2. Find critical numbers: $g'(y)=0$ when $5y+12=0 \implies y = -12/5 = -2.4$. $g'(y)$ is undefined when $y+4=0 \implies y=-4$. Both are in the interval $[-5, 1]$.

3. Evaluate the function at critical numbers and endpoints:

• $g(-5) = 3(-5)(-5+4)^{2/3} = -15(-1)^{2/3} = -15$
• $g(-4) = 3(-4)(-4+4)^{2/3} = 0$
• $g(-12/5) = 3(-\frac{12}{5})(-\frac{12}{5}+4)^{2/3} = -\frac{36}{5}(\frac{8}{5})^{2/3} = -\frac{36}{5}\frac{4}{5^{2/3}} = -\frac{144}{5^{5/3}} \approx -15.06$
• $g(1) = 3(1)(1+4)^{2/3} = 3(5)^{2/3} \approx 8.77$

4. Compare the values: The largest value is $3 \cdot 5^{2/3}$ and the smallest is $-144/5^{5/3}$.

Since the exponential function $e^u$ is always increasing, the maximum and minimum values of $f(x)=e^{g(x)}$ will occur at the same $x$-values that maximize and minimize the exponent function, $g(x) = x^3-2x^2-7x$. Find the extrema of the exponent on the given interval.

The absolute maximum is $e^{23/8}$, which occurs at $x = -1/2$.
The absolute minimum is $e^{-392/27}$, which occurs at $x = 7/3$.

Let $g(x) = x^3-2x^2-7x$. We will find the extrema of $g(x)$ on $[-\frac{1}{2}, \frac{5}{2}]$.

1. Find the derivative of the exponent: $$ g'(x) = 3x^2 - 4x - 7 $$ 2. Find critical numbers: Set $g'(x)=0$. $$ 3x^2 - 4x - 7 = 0 \implies (3x-7)(x+1) = 0 $$ The critical numbers are $x=7/3$ and $x=-1$. Only $x=7/3$ is within the interval $[-\frac{1}{2}, \frac{5}{2}]$.

3. Evaluate $g(x)$ at the critical number and endpoints:

• $g(-\frac{1}{2}) = (-\frac{1}{2})^3 - 2(-\frac{1}{2})^2 - 7(-\frac{1}{2}) = -\frac{1}{8} - \frac{2}{4} + \frac{7}{2} = \frac{-1-4+28}{8} = \frac{23}{8}$
• $g(\frac{7}{3}) = (\frac{7}{3})^3 - 2(\frac{7}{3})^2 - 7(\frac{7}{3}) = \frac{343}{27} - \frac{98}{9} - \frac{49}{3} = \frac{343 - 294 - 441}{27} = -\frac{392}{27}$
• $g(\frac{5}{2}) = (\frac{5}{2})^3 - 2(\frac{5}{2})^2 - 7(\frac{5}{2}) = \frac{125}{8} - \frac{50}{4} - \frac{35}{2} = \frac{125 - 100 - 140}{8} = -\frac{115}{8}$

4. Compare the values of $g(x)$: The maximum value of the exponent is $23/8$ and the minimum is $-392/27$. The maximum and minimum of $f(x)$ are therefore $e^{23/8}$ and $e^{-392/27}$.

This is an optimization problem. The domain for the number of clinics is $x>0$. Find the derivative $C'(x)$, set it to zero, and solve for $x$ to find the critical number. Since the number of clinics must be an integer, test the integers on either side of your critical number to find the true minimum.

Either 4 or 5 clinics will minimize the cost.

1. Find the derivative of the cost function: $$ C(x) = 500x + 10000x^{-1} $$ $$ C'(x) = 500 - 10000x^{-2} = 500 - \frac{10000}{x^2} $$ 2. Find the critical number: Set $C'(x)=0$. $$ 500 = \frac{10000}{x^2} \implies x^2 = \frac{10000}{500} = 20 $$ $$ x = \sqrt{20} \approx 4.47 $$ 3. Consider the integer domain: Since the number of clinics must be an integer, the minimum must occur at one of the integers closest to the critical value of $4.47$. We need to test $x=4$ and $x=5$.

• $C(4) = 500(4) + \frac{10000}{4} = 2000 + 2500 = \$4500$
• $C(5) = 500(5) + \frac{10000}{5} = 2500 + 2000 = \$4500$

Both 4 and 5 clinics result in the same minimum cost of \$4500. The second derivative, $C''(x) = \frac{20000}{x^3}$, is positive for $x > 0$, confirming that the function is concave up and our critical point is a minimum.

This is a maximization problem common in statistics (Maximum Likelihood Estimation). Treat everything except $p$ as a constant. Differentiate $\ell$ with respect to $p$, set the result to zero, and solve for $p$. It's helpful to first abbreviate $\sum_{i=1}^n x_i$ as a constant, like $S$.

$$ p = \frac{nr}{\sum_{i=1}^n x_i + nr} $$

Let the constant $S = \sum_{i=1}^n x_i$. The function is $\ell(p) = S \cdot \log(1-p) + nr \cdot \log(p)$. The domain for a probability $p$ is $(0,1)$.

1. Differentiate with respect to $p$: $$ \frac{d\ell}{dp} = S \cdot \frac{d}{dp}(\log(1-p)) + nr \cdot \frac{d}{dp}(\log(p)) $$ Using the chain rule for the first term: $$ \frac{d\ell}{dp} = S \left(\frac{1}{1-p} \cdot (-1)\right) + nr \left(\frac{1}{p}\right) = -\frac{S}{1-p} + \frac{nr}{p} $$ 2. Set the derivative to zero and solve for $p$: $$ \frac{nr}{p} = \frac{S}{1-p} $$ $$ nr(1-p) = S \cdot p $$ $$ nr - nr \cdot p = S \cdot p $$ $$ nr = S \cdot p + nr \cdot p = p(S + nr) $$ $$ p = \frac{nr}{S + nr} $$ Substituting $S$ back gives the final answer: $$ p = \frac{nr}{\sum_{i=1}^n x_i + nr} $$ The second derivative is $\frac{d^2\ell}{dp^2} = -\frac{S}{(1-p)^2} - \frac{nr}{p^2}$, which is negative for typical parameter values, confirming this is a maximum.

Treat $n$ and the summation $\sum \log(x_i)$ as known constants. Differentiate the entire expression with respect to $\theta$, set the result to zero, and solve.

$$ \theta = -\frac{n}{\sum_{i=1}^n \log(x_i)} $$

Let the constant $L = \sum_{i=1}^n \log(x_i)$. The function can be rewritten as: $$ \ell(\theta) = n\log(\theta) + (\theta-1)L = n\log(\theta) + \theta L - L $$ 1. Differentiate with respect to $\theta$: $$ \frac{d\ell}{d\theta} = \frac{d}{d\theta}(n\log(\theta)) + \frac{d}{d\theta}(\theta L) - \frac{d}{d\theta}(L) $$ $$ \frac{d\ell}{d\theta} = \frac{n}{\theta} + L - 0 $$ 2. Set the derivative to zero and solve for $\theta$: $$ \frac{n}{\theta} + L = 0 \implies \frac{n}{\theta} = -L $$ $$ \theta = -\frac{n}{L} $$ Substituting $L$ back gives the final answer: $$ \theta = -\frac{n}{\sum_{i=1}^n \log(x_i)} $$ The second derivative, $\frac{d^2\ell}{d\theta^2} = -\frac{n}{\theta^2}$, is negative (assuming $n>0$), which confirms this value of $\theta$ yields a maximum.