To determine the relative extrema, we first need to determine its derivative.

\(y'\) = \(\Large{\frac{d}{dx}}\)\([x^4 - x^3]\) = \(4x^3 - 3x^2\) = \(x^2(4x - 3)\)

We can infer \(y'\) is continuous at all points and is zero (0) at two points: x = 0 and \(x = \Large{\frac{3}{4}}\). These are the critical numbers of the given function y.

The following graph is the illustration of the functions y (in green) and with the critical numbers as red dots:

Fig.1

The critical numbers of the function f(x) partition the domain of the function f(x) into intervals \((-\infty, 0)\), \((0, \Large{\frac{3}{4}}\normalsize{)}\), and \((\Large{\frac{3}{4}}, \normalsize{\infty})\) respectively.

Now it is time to find the sign of the derivative function \(y'\) in each of the 3 domain intervals by picking a test number in each of the domain intervals.

RESULT: applying the first-derivative test to the above table, there is no relative extrema at x = 0 since the derivative is negative on the left and right. At x = \(\Large{\frac{3}{4}}\), we have the derivative as negative to the left and positive to the right and therefore we have a relative minimum.

To determine the relative extrema, we first need to determine its derivative.

\(y'\) = \(\Large{\frac{d}{dx}}\)\([2x - 3x^{\frac{2}{3}}]\) = \(4 - 3.\Large{\frac{2}{3}}\).\(x^{\frac{2}{3}-1}\) = \(2(1 - \Large{\frac{1}{x^{\frac{1}{3}}}}\)\()\) = \(2\Large{\frac{(x^{\frac{1}{3}} - 1)}{x^{\frac{1}{3}}}}\)

We can infer \(y'\) is continuous at all points and (a) is undefined at zero (0) and (b) zero (0) at x = 1. These are the critical numbers of the given function y.

The following graph is the illustration of the functions y (in green) and with the critical numbers as red dots:

Fig.2

The critical numbers of the function f(x) partition the domain of the function f(x) into intervals \((-\infty, 0)\), \((0, 1)\), and \((1, \infty)\) respectively.

Now it is time to find the sign of the derivative function \(y'\) in each of the 3 domain intervals by picking a test number in each of the domain intervals.

RESULT: applying the first-derivative test to the above table, we have a relative maximum at x = 0 since the derivative is negative on the left and positive on the right. Similarly, at x = 1, we have a relative minimum since the derivative is positive to the left and negative to the right.

Let x be the first number and y be the second number. Given xy = 288. We want to minimize the sum S = 2x + y.

Rewriting the equation for S to depend on one variable (say x), \(S = 2x + \Large{\frac{288}{x}}\).

Given that the numbers are positive, the domain if \(x \geq 0\).

Next, we find the critical number(s) to help us identify the minimum value of S. To find the critical number(s), we need to find the derivative of S. That is \(S'\) = \(\Large{\frac{d}{dx}}\)\([2x - \Large{\frac{288}{x}}\)\(]\) = \(2 - \Large{\frac{288}{x^2}}\)

To find the critical number(s), we set the derivative \(S' = 2 - \Large{\frac{288}{x^2}}\) = \(0\).

That is, \(x^2 = 144\) or \(x = \pm12\). Since the domain is only positive numbers, we choose x = 12.

The critical number x = 12 partitions the domain of the function S into intervals \((0, 12)\) and \((12, \infty)\) respectively.

Now it is time to find the sign of the derivative function \(S'\) in each of the 2 domain intervals by picking a test number in each of the domain intervals.

RESULT: applying the first-derivative test to the above table, we have a relative minimum at x = 12, since the derivative is negative to the left and positive to the right. Therefore, the two numbers are x = 12 and y = 24.

The given equation for the function is \(f(x) = \Large{\frac{3}{x^2 + 3}}\).

Given \(f(x) = \Large{\frac{3}{x^2 + 3}}\), the first derivative of \(f(x) = 6.(x^2 + 3)^{-1}\) using the chain rule is \(f'(x) = 6.(-1).(2x).(x^2 + 3)^{-2}\) = \(\Large{\frac{-12x}{(x^2 + 3)^2}}\).

The second derivative \(f''(x)\) can be found using the quotient rule. Therefore, \(f''(x) = \Large{\frac{(x^2 + 3)^2(-12) - (-12x)(2)(2x)(x^2 +3)}{(x^2 + 3)^4}}\) = \(\Large{\frac{-12(x^2 + 3) + 48x^2}{(x^2 + 3)^3}}\) = \(\Large{\frac{36(x^2 - 1)}{(x^2 + 3)^3}}\).

From the above equation for \(f''(x)\), we can infer the second derivation is defined for all real numbers. Also, for x = -1 or 1, \(f''(x) = 0\).

Therefore, the domain of the function \(f''(x)\) can be divided into intervals \((-\infty, -1)\), \(-1, 1\), and \((1, \infty)\) respectively.

Now it is time to find if the second derivative function \(f''(x)\) is increasing or decreasing in each of the 3 domain intervals by picking a test number in each of the intervals.

The following illustration shows the conclusion on the concavity of the function f(x):

Fig.3

(a) The weight of the culture after \(t = 10\) hours, \(y = \Large{\frac{1.25}{1 + 0.25e^{-0.4(10)}}}\)

RESULT: The weight \(y \approx 1.244\) grams

(b) The limit of the model as t increases without bound (to \(\infty\)) is \(\lim_{t \to \infty} \Large{\frac{1.25}{1 + 0.25e^{-0.4t}}}\) = \(\lim_{t \to \infty} \Large{\frac{1.25}{1 + \frac{0.25}{e^{0.4t}}}}\)

As \(t \rightarrow \infty\), the term \(e^{0.4t}\) gets very large. This implies \(\Large{\frac{0.25}{e^{0.4t}}}\) will approach zero (0). Therefore, \(y = \Large{\frac{1.25}{1 + 0}}\) = 1.25

RESULT: The weight of the culture approaches 1.25 as t increases without bound.

The balance amount A as n (number of interest compound) increases without bound (to \(\infty\)) is \(\lim_{n \to \infty} P\Large{(1 + \frac{r}{n})^n}\) = \(\lim_{n \to \infty} P\Large{[(1 + \frac{r}{n})^{\frac{n}{r}}]^r}\)

Let \(x = \Large{\frac{r}{n}}\), then \(\lim_{x \to 0} P(1 + x)^{\frac{1}{x}}\) = \(P[\lim_{x \to 0}(1 + x)^{\frac{1}{x}}]\)

We know \(\lim_{x \to 0} (1 + x)^{\Large{\frac{1}{x}}}\)\( = e\).

RESULT: The limit of the balanace A as n increases without bound is \(lim_{n \to \infty} A = Pe^r\).

We need to use the chain rule to find the derivative of f(x). Let \(u = -3x^2\).

Then, \(f(x) = e^{-3x^2} = e^u\). We know \(\Large{\frac{d}{dx}}\)\([e^u]\) = \(e^u.\Large{\frac{du}{dx}}\).

Therefore, \(f'(x) = e^u.\Large{\frac{du}{dx}}\) = \(e^u.(-3)(2)x^{2 - 1} = -6x.e^u = -6xe^{-3x^2}\)

RESULT: The derivative of f(x) is \(-6xe^{-3x^2}\).

The following diagram provides a pictorial illustration:

Fig.6

Given the domain of x is the closed interval \([-30, 30]\).

To find the lowest (or minimum) point on the telephone wire, we need to find the first-derivative of the function and find the critical number.

Let \(u = \frac{x}{60}\). Therefore, \(y' = \Large{\frac{d}{dx}}\)[\(30(e^u + e^{-u})\)] = \(30[e^u.\frac{du}{dx} + e^{-u}.\frac{du}{dx}]\) = \(30[e^u.(\frac{1}{60}) + e^{-u}.(-\frac{1}{60})]\) = \(30[\frac{e^u}{60} - \frac{e^{-u}}{60}]\) = \(\frac{1}{2}(e^u - e^{-u})\) = \(\frac{1}{2}(e^{\frac{x}{60}} - e^{-\frac{x}{60}})\)

To find the critical number, we set \(y' = 0\). That is, \(\frac{1}{2}(e^{\frac{x}{60}} - e^{-\frac{x}{60}}) = 0\)

That is \(e^{\frac{x}{60}} - e^{-\frac{x}{60}} = 0\). OR \(\frac{x}{60} = -\frac{x}{60}\). OR \(x = -x\). OR \(2x = 0\). This implies \(x = 0\).

The height the telephone wire (from the ground) can be computed by substituting \(x = -30\), \(x = 0\), and \(x = 30\) in the function for \(y\) to figure the lowest (minimum) point. The following are the three values:

RESULT: The height from the ground at the lowest point in the telephone wire is 60 feet.