Practice Final #3

We have a function with two terms:
1. $x^3e^{2x}$
2. $3(x - 2)^2e^{-x}$

We'll need to find the derivative of each term using the product rule, then evaluate at $x = 0$.



For the first term, $x^3e^{2x}$, we use the product rule:

$$\begin{align*}$$
& \frac{d}{dx}[x^3e^{2x}] \\
&= \frac{d}{dx}[x^3] \cdot e^{2x} + x^3 \cdot \frac{d}{dx}[e^{2x}] \\
&= 3x^2 \cdot e^{2x} + x^3 \cdot 2e^{2x} \\
&= 3x^2e^{2x} + 2x^3e^{2x} \\
&= e^{2x}(3x^2 + 2x^3)
\end{align*}



For the second term, $3(x - 2)^2e^{-x}$, we use the product rule again:


$$\begin{align*}$$
& \frac{d}{dx}[3(x - 2)^2e^{-x}] \\
&= \frac{d}{dx}[3(x - 2)^2] \cdot e^{-x} + 3(x - 2)^2 \cdot \frac{d}{dx}[e^{-x}] \\
&= 3 \cdot 2(x - 2) \cdot e^{-x} + 3(x - 2)^2 \cdot (-1)e^{-x} \\
&= 6(x - 2)e^{-x} - 3(x - 2)^2e^{-x} \\
&= e^{-x}[6(x - 2) - 3(x - 2)^2]
\end{align*}


When differentiating terms with $e^{ax}$, remember that $\frac{d}{dx}[e^{ax}] = a \cdot e^{ax}$. The coefficient comes out front!



The complete derivative is:


$$\begin{align*}$$
& f'(x) \\
&= e^{2x}(3x^2 + 2x^3) \\
&+ e^{-x}[6(x - 2) - 3(x - 2)^2]
\end{align*}

Now we evaluate at $x = 0$:

$$\begin{align*}$$
& f'(0) \\
&= e^{2(0)}(3(0)^2 + 2(0)^3) \\
&+ e^{-(0)}[6(0 - 2) - 3(0 - 2)^2] \\
&= e^0(0) \\
&+ e^0[6(-2) - 3(-2)^2] \\
&= 1 \cdot 0 \\
&+ 1 \cdot [(-12) - 3 \cdot 4] \\
&= 0 \\
&+ [-12 - 12] \\
&= -24
\end{align*}
Therefore:

$$\boxed{f'(0) = -24}$$

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

As $x \to \infty$:
- $\sin(x)/e^x \to 0$ since $e^x$ grows much faster than $\sin(x)$
- $\cos(x)/e^x \to 0$ for the same reason

When dealing with limits involving exponentials and trigonometric functions, remember that exponential functions like $e^x$ grow much faster than any polynomial or trigonometric function. Since sine and cosine are always bounded between -1 and 1, the ratio $\sin(x)/e^x$ or $\cos(x)/e^x$ will always approach 0 as $x$ approaches infinity.





Therefore:
$$\boxed{\lim_{x \to \infty} \frac{3e^x + 5\sin(x)}{e^x + \cos(x)} = 3}$$

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.

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