Practice Final #2

For a piecewise function to be continuous at the transition point (in this case $x = 0$), the limits from both sides must be equal:

$$\begin{align*}$$
& \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)
\end{align*}



The left-hand limit uses the first piece of the function:

$$\begin{align*}$$
& \lim_{x \to 0^-} f(x) \\
&= \lim_{x \to 0^-} \sinh(x) \\
&= \sinh(0) \\
&= 0
\end{align*}



The right-hand limit uses the second piece:

$$\begin{align*}$$
& \lim_{x \to 0^+} f(x) \\
&= \lim_{x \to 0^+} (b - 3\cos(x)) \\
&= b - 3\cos(0) \\
&= b - 3(1) \\
&= b - 3
\end{align*}



For continuity, these limits must be equal:

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)$$
$$\sinh(0) = b - 3\cos(0)$$
$$0 = b - 3(1)$$
$$0 = b - 3$$
$$b = 3$$

When working with piecewise functions, always check the value of each piece at the transition point to ensure continuity.

Therefore, the value of $b$ that makes $f(x)$ continuous everywhere is:

$$\boxed{b = 3}$$

We need to differentiate both sides of the equation with respect to $x$:
$$\begin{align*}$$
\tanh(x) &= xy
\end{align*}

Differentiating the left side:
$$\begin{align*}$$
\frac{d}{dx}[\tanh(x)] &= \text{sech}^2(x)
\end{align*}

Differentiating the right side:
$$\begin{align*}$$
\frac{d}{dx}[xy] &= y + x\frac{dy}{dx}
\end{align*}



$$\begin{align*}$$
\text{sech}^2(x) &= y + x\frac{dy}{dx} \\
\text{sech}^2(x) - y &= x\frac{dy}{dx} \\
\frac{dy}{dx} &= \frac{\text{sech}^2(x) - y}{x}
\end{align*}



At the point $(1, \tanh(1))$, we have $x = 1$ and $y = \tanh(1)$.

Let's substitute these values:
$$\begin{align*}$$
\frac{dy}{dx}\bigg|_{(1,\tanh(1))} &= \frac{\text{sech}^2(1) - \tanh(1)}{1} \\
&= \text{sech}^2(1) - \tanh(1)
\end{align*}

Now we need to calculate these values:
$$\begin{align*}$$
\text{sech}^2(1) &= \frac{1}{\cosh^2(1)} \\
&= \frac{1}{(\frac{e + e^{-1}}{2})^2} \\
&= \frac{4}{(e + e^{-1})^2}
\end{align*}

And:
$$\begin{align*}$$
\tanh(1) &= \frac{\sinh(1)}{\cosh(1)} \\
&= \frac{\frac{e - e^{-1}}{2}}{\frac{e + e^{-1}}{2}} \\
&= \frac{e - e^{-1}}{e + e^{-1}}
\end{align*}

Therefore:

$$\begin{align*}$$
\frac{dy}{dx} &= \frac{4}{(e + e^{-1})^2} - \frac{e - e^{-1}}{e + e^{-1}} \\
&= \frac{4}{(e + e^{-1})^2} - \frac{(e - e^{-1})(e + e^{-1})}{(e + e^{-1})^2} \\
&= \frac{4 - (e^2 - e^{-2})}{(e + e^{-1})^2} \\
&= \frac{4 - (e^2 - e^{-2})}{(e + e^{-1})^2}
\end{align*}


We can simplify further:
$$\begin{align*}$$
\frac{dy}{dx} &= \frac{4 - e^2 + e^{-2}}{(e + e^{-1})^2}
\end{align*}

Therefore:
$$\boxed{\frac{dy}{dx} = \frac{4 - e^2 + e^{-2}}{(e + e^{-1})^2}}$$

When differentiating hyperbolic functions, remember that $\frac{d}{dx}[\tanh(x)] = \text{sech}^2(x) = \frac{1}{\cosh^2(x)}$, similar to how the derivative of $\tan(x)$ is $\sec^2(x)$ for trigonometric functions.

Let $f(x) = 3x^{\frac{2}{3}} - 5x^{-\frac{3}{4}}$. Find $f'(1)$.



We need to use the power rule for differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$

For the first term: $3x^{\frac{2}{3}}$
$$\begin{align*}$$
\frac{d}{dx}(3x^{\frac{2}{3}}) &= 3 \cdot \frac{2}{3} \cdot x^{\frac{2}{3}-1} \\
&= 2 \cdot x^{-\frac{1}{3}} \\
&= \frac{2}{x^{\frac{1}{3}}}
\end{align*}

For the second term: $-5x^{-\frac{3}{4}}$
$$\begin{align*}$$
\frac{d}{dx}(-5x^{-\frac{3}{4}}) &= -5 \cdot (-\frac{3}{4}) \cdot x^{-\frac{3}{4}-1} \\
&= -5 \cdot (-\frac{3}{4}) \cdot x^{-\frac{7}{4}} \\
&= \frac{15}{4} \cdot x^{-\frac{7}{4}} \\
&= \frac{15}{4 \cdot x^{\frac{7}{4}}}
\end{align*}

Now we combine these results to find $f'(x)$:
$$\begin{align*}$$
f'(x) &= \frac{2}{x^{\frac{1}{3}}} + \frac{15}{4 \cdot x^{\frac{7}{4}}}
\end{align*}



When we substitute $x = 1$:
$$\begin{align*}$$
f'(1) &= \frac{2}{1^{\frac{1}{3}}} + \frac{15}{4 \cdot 1^{\frac{7}{4}}} \\
&= \frac{2}{1} + \frac{15}{4 \cdot 1} \\
&= 2 + \frac{15}{4} \\
&= 2 + \frac{15}{4} \\
&= \frac{8}{4} + \frac{15}{4} \\
&= \frac{23}{4}
\end{align*}

Therefore:
$$\boxed{f'(1) = \frac{23}{4}}$$

I understand now. Let me correct the formatting with proper math brackets and vertical alignment:


Evaluate the following limit:
$$\lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h}$$








$$\begin{align*}$$
\lim_{h \to 0} (-8 + h) &= -8 + 0 \\
&= -8
\end{align*}

Therefore:
$$\boxed{\lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h} = -8}$$

$$\lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6}$$

For the numerator:
$$t^2 - 9 = (t-3)(t+3)$$

For the denominator:




$$\lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6}$$
$$= \lim_{t \to -3} \frac{(t-3)(t+3)}{(3t+2)(t+3)}$$
$$= \lim_{t \to -3} \frac{(t-3)\cancel{(t+3)}}{(3t+2)\cancel{(t+3)}}$$
$$= \lim_{t \to -3} \frac{t-3}{3t+2}$$



$$\lim_{t \to -3} \frac{t-3}{3t+2}$$
$$= \frac{-3-3}{3(-3)+2}$$
$$= \frac{-6}{-9+2}$$
$$= \frac{-6}{-7}$$
$$= \frac{6}{7}$$

Therefore:
$$\boxed{\lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6} = \frac{6}{7}}$$

Let me revise the solution to explicitly find the critical points and exclude them from the interval:

Let $f(x) = |x|^{\frac{4}{3}}(9 - x)$. Find the set on which the function is increasing.



When $x > 0$, $|x| = x$, so:
$$\begin{align*}$$
& f(x) = x^{\frac{4}{3}}(9 - x) \\
& f'(x) = \frac{4}{3}x^{\frac{1}{3}}(9 - x) - x^{\frac{4}{3}} \\
& f'(x) = x^{\frac{1}{3}}[\frac{4}{3}(9 - x) - x] \\
& f'(x) = x^{\frac{1}{3}}[12 - \frac{7}{3}x]
\end{align*}



When $x < 0$, $|x| = -x$, so:
$$\begin{align*}$$
& f(x) = (-x)^{\frac{4}{3}}(9 - x) \\
& f'(x) = (-x)^{\frac{1}{3}}[-12 + \frac{1}{3}x]
\end{align*}



Critical points occur when $f'(x) = 0$ or when $f'(x)$ is undefined.

For $x > 0$:
$$\begin{align*}$$
& f'(x) = x^{\frac{1}{3}}[12 - \frac{7}{3}x] = 0
\end{align*}

This occurs when:
- $x^{\frac{1}{3}} = 0$, which is impossible for $x > 0$
- $12 - \frac{7}{3}x = 0$, which gives $x = \frac{36}{7}$

For $x < 0$:
$$\begin{align*}$$
& f'(x) = (-x)^{\frac{1}{3}}[-12 + \frac{1}{3}x] = 0
\end{align*}

This occurs when:
- $(-x)^{\frac{1}{3}} = 0$, which is impossible for $x < 0$
- $-12 + \frac{1}{3}x = 0$, which gives $x = 36$, but this is outside our domain of $x < 0$

For $x = 0$:
Both formulations of the derivative approach 0 as $x$ approaches 0, so $x = 0$ is also a critical point.

Therefore, our critical points are $x = 0$ and $x = \frac{36}{7}$.



For $x < 0$:
$$f'(x) = (-x)^{\frac{1}{3}}[-12 + \frac{1}{3}x]$$

Since $(-x)^{\frac{1}{3}} > 0$ for $x < 0$, and $[-12 + \frac{1}{3}x] < 0$ for $x < 0$ (because $\frac{1}{3}x < 0$ and $-12 < 0$), we have $f'(x) < 0$ for all $x < 0$.

For $0 < x < \frac{36}{7}$:
$$f'(x) = x^{\frac{1}{3}}[12 - \frac{7}{3}x]$$

Since $x^{\frac{1}{3}} > 0$ for $x > 0$, and $[12 - \frac{7}{3}x] > 0$ for $x < \frac{36}{7}$, we have $f'(x) > 0$ for $0 < x < \frac{36}{7}$.

For $x > \frac{36}{7}$:
$$f'(x) = x^{\frac{1}{3}}[12 - \frac{7}{3}x]$$

Since $x^{\frac{1}{3}} > 0$ for $x > 0$, and $[12 - \frac{7}{3}x] < 0$ for $x > \frac{36}{7}$, we have $f'(x) < 0$ for $x > \frac{36}{7}$.



Based on our analysis:
- For $x < 0$: $f'(x) < 0$ (function is decreasing)
- For $0 < x < \frac{36}{7}$: $f'(x) > 0$ (function is increasing)
- For $x > \frac{36}{7}$: $f'(x) < 0$ (function is decreasing)
- At $x = 0$ and $x = \frac{36}{7}$: $f'(x) = 0$ (function is neither increasing nor decreasing)

Since we're looking for where the function is strictly increasing (not just non-decreasing), we exclude the critical points.

Therefore, the set on which $f(x) = |x|^{\frac{4}{3}}(9 - x)$ is increasing is:

$$\boxed{(0, \frac{36}{7})}$$

When finding intervals of increase/decrease, exclude critical points where the derivative equals zero, as the function is not strictly increasing or decreasing at those points.

Let $u(x) = (x - 1)^2$ and $v(x) = x + 2$

$$\begin{align*}$$
& \frac{dy}{dx} \\
&= \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}
\end{align*}

Finding the derivatives:

$$\begin{align*}$$
& u'(x) \\
&= \frac{d}{dx}[(x - 1)^2] \\
&= 2(x - 1)
\end{align*}

$$\begin{align*}$$
& v'(x) \\
&= \frac{d}{dx}[x + 2] \\
&= 1
\end{align*}

Substituting into the quotient rule:

$$\begin{align*}$$
& \frac{dy}{dx} \\
&= \frac{2(x - 1)(x + 2) - (x - 1)^2 \cdot 1}{(x + 2)^2}
\end{align*}



At $x = 2$:

$$\begin{align*}$$
& \frac{dy}{dx}\bigg|_{x=2} \\
&= \frac{2(2 - 1)(2 + 2) - (2 - 1)^2}{(2 + 2)^2} \\
&= \frac{2(1)(4) - (1)^2}{4^2} \\
&= \frac{8 - 1}{16} \\
&= \frac{7}{16}
\end{align*}



The slope of the tangent line at $(2, 1)$ is $m = \frac{7}{16}$

Using the point-slope form of a line:

$$\begin{align*}$$
& y - y_1 = m(x - x_1) \\
&y - 1 = \frac{7}{16}(x - 2) \\
&y - 1 = \frac{7}{16}x - \frac{7}{8} \\
&y = \frac{7}{16}x - \frac{7}{8} + 1 \\
&y = \frac{7}{16}x - \frac{7}{8} + \frac{8}{8} \\
&y = \frac{7}{16}x + \frac{1}{8}
\end{align*}



Let's find a point with a simple x-value, say $x = 0$:

$$\begin{align*}$$
& y = \frac{7}{16} \cdot 0 + \frac{1}{8} \\
&y = 0 + \frac{1}{8} \\
&y = \frac{1}{8}
\end{align*}

Therefore, the point $(0, \frac{1}{8})$ lies on the tangent line.

$$\boxed{(x,y) = \left(0, \frac{1}{8}\right)}$$

For a constant $b > 0$, consider the plane region bounded by the curves $y = bx(1-x)$ and $y = b\sin(\pi x)$ in the first quadrant.



The curves intersect when:
$$bx(1-x) = b\sin(\pi x)$$
$$x(1-x) = \sin(\pi x)$$

The solutions are $x = 0$ and $x = 1$ since:
- At $x = 0$: $0 \cdot (1-0) = \sin(0) \Rightarrow 0 = 0$ ✓
- At $x = 1$: $1 \cdot (1-1) = \sin(\pi) \Rightarrow 0 = 0$ ✓

For problems with intersecting curves, identify where they cross and determine which function has larger values in each subinterval to properly set up the volume integrals.



Let's compare $f(x) = x(1-x)$ and $g(x) = \sin(\pi x)$ in the interval $[0,1]$.

Both functions reach their maximum values at $x = 0.5$:
- $f(0.5) = 0.5 \cdot 0.5 = 0.25$
- $g(0.5) = \sin(\pi/2) = 1$

Since $g(0.5) > f(0.5)$, the sine function is always above the parabola in the interval $(0,1)$.



When rotating around the x-axis, we use the washer method where:
- Outer radius $R = b\sin(\pi x)$ (sine function)
- Inner radius $r = bx(1-x)$ (parabola)

$$V_x = \pi\int_{0}^{1} [(b\sin(\pi x))^2 - (bx(1-x))^2] dx$$

$$V_x = \pi b^2\int_{0}^{1} [\sin^2(\pi x) - x^2(1-x)^2] dx$$

For the first term, using $\sin^2(\pi x) = \frac{1-\cos(2\pi x)}{2}$:
$$\int_{0}^{1} \sin^2(\pi x) dx = \int_{0}^{1} \frac{1-\cos(2\pi x)}{2} dx$$

$$= \frac{1}{2}\int_{0}^{1} dx - \frac{1}{2}\int_{0}^{1} \cos(2\pi x) dx$$

$$= \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{2\pi}[\sin(2\pi x)]_{0}^{1}$$

$$= \frac{1}{2} - 0 = \frac{1}{2}$$

For the second term:
$$\int_{0}^{1} x^2(1-x)^2 dx = \int_{0}^{1} x^2(1-2x+x^2) dx$$

$$= \int_{0}^{1} (x^2 - 2x^3 + x^4) dx$$

$$= \left[\frac{x^3}{3} - \frac{2x^4}{4} + \frac{x^5}{5}\right]_{0}^{1}$$

$$= \frac{1}{3} - \frac{1}{2} + \frac{1}{5} = \frac{10-15+6}{30} = \frac{1}{30}$$

Therefore:
$$V_x = \pi b^2\left(\frac{1}{2} - \frac{1}{30}\right)$$

$$V_x = \pi b^2 \cdot \frac{15-1}{30}$$

$$V_x = \pi b^2 \cdot \frac{14}{30} = \pi b^2 \cdot \frac{7}{15}$$



When rotating around the y-axis, we use the cylindrical shell method.

For each x from 0 to 1:
- Shell circumference = 2$\pi x$
- Shell height = $b\sin(\pi x) - bx(1-x)$ (difference between the two curves)
- Shell thickness = dx

$$V_y = 2\pi\int_{0}^{1} x \cdot [b\sin(\pi x) - bx(1-x)] dx$$

$$V_y = 2\pi b\int_{0}^{1} [x\sin(\pi x) - x^2(1-x)] dx$$

$$V_y = 2\pi b\int_{0}^{1} [x\sin(\pi x) - x^2 + x^3] dx$$

For the first term, using integration by parts:
$$\int_{0}^{1} x\sin(\pi x) dx = \left[-\frac{x\cos(\pi x)}{\pi}\right]_{0}^{1} + \frac{1}{\pi}\int_{0}^{1} \cos(\pi x) dx$$

$$= -\frac{1\cdot(-1)}{\pi} - 0 + \frac{1}{\pi} \cdot \frac{1}{\pi}[\sin(\pi x)]_{0}^{1}$$

$$= \frac{1}{\pi} + 0 = \frac{1}{\pi}$$

For the second term:
$$\int_{0}^{1} -x^2 dx = -\frac{x^3}{3}\bigg|_{0}^{1} = -\frac{1}{3}$$

For the third term:
$$\int_{0}^{1} x^3 dx = \frac{x^4}{4}\bigg|_{0}^{1} = \frac{1}{4}$$

Therefore:
$$V_y = 2\pi b\left(\frac{1}{\pi} - \frac{1}{3} + \frac{1}{4}\right)$$

$$V_y = 2\pi b\left(\frac{1}{\pi} + \frac{3-4}{12}\right)$$

$$V_y = 2\pi b\left(\frac{1}{\pi} - \frac{1}{12}\right)$$



We want:
$$V_x\left(\frac{1}{\pi} - \frac{1}{12}\right) = V_y$$

$$\pi b^2 \cdot \frac{7}{15} \cdot \left(\frac{1}{\pi} - \frac{1}{12}\right) = 2\pi b\left(\frac{1}{\pi} - \frac{1}{12}\right)$$

The term $\left(\frac{1}{\pi} - \frac{1}{12}\right)$ appears on both sides, so we can divide by it:

$$\pi b^2 \cdot \frac{7}{15} = 2\pi b$$

$$\frac{7b^2}{15} = 2b$$

$$7b^2 = 30b$$

$$7b = 30$$

$$b = \frac{30}{7}$$

Therefore:
$$\boxed{b = \frac{30}{7}}$$