### **MATH 150A - Calculus I: Practice Materials**
#### **Chapter 1: Limits and Continuity**
1. **Evaluating Limits Algebraically:**
Find the value of the limit:
$$\lim_{x \to -3} \frac{x^2 + x - 6}{x + 3}$$
2. **Limits at Infinity:**
Determine the limit:
$$\lim_{x \to \infty} \frac{5x^3 - 2x^2 + 1}{10x^3 + 3x - 7}$$
3. **Continuity of Piecewise Functions:**
Find the value of the constant $c$ that makes the function $f(x)$ continuous on $(-\infty, \
infty)$.
$$f(x) = \begin{cases} cx^2 + 2x & \text{if } x < 2 \\ x^3 - cx & \text{if } x \ge 2 \end{cases}$$
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#### **Chapter 2: Derivatives**
1. **Derivative from the Limit Definition:**
Using the limit definition of the derivative, $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, find the
derivative of $f(x) = x^2 - 4x$.
2. **Applying Differentiation Rules:**
Find the derivative, $\frac{dy}{dx}$, for the function $y = e^{x^2} \sin(3x)$. Do not simplify.
3. **Implicit Differentiation:**
Find $\frac{dy}{dx}$ for the curve defined by the equation $y^2 + x^2y = 5x$.
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#### **Chapter 3: Applications of Differentiation**
1. **Analyzing Functions (First Derivative Test):**
For the function $f(x) = 2x^3 + 3x^2 - 12x + 1$:
* Find the critical numbers.
* Find the intervals on which the function is increasing or decreasing.
* Identify the local maximum and minimum values.
2. **Concavity and Inflection Points (Second Derivative Test):**
For the function $g(x) = x^4 - 6x^2$:
* Find the intervals of concavity (where the function is concave upward or downward).
* Find the inflection points.
3. **Optimization:**
A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide
it in half with a fence parallel to one of the sides of the rectangle. What should the dimensions of
the field be to minimize the cost of the fence?
4. **Related Rates:**
A cylindrical tank with a radius of 5 meters is being filled with water at a rate of 3 cubic meters
per minute. How fast is the height of the water increasing? (The volume of a cylinder is $V = \pi
r^2 h$).
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#### **Chapter 4: Integrals**
1. **The Fundamental Theorem of Calculus:**
Evaluate the definite integral:
$$\int_0^{\pi/2} (\cos(x) + 3) \,dx$$
2. **Indefinite Integrals with U-Substitution:**
Find the indefinite integral:
$$\int \frac{x}{(x^2+1)^2} \,dx$$
3. **Area Between Curves:**
Find the area of the region enclosed by the parabolas $y = x^2$ and $y = 2x - x^2$.
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### **Solutions**
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#### **Chapter 1 Solutions**
1. **Solution:** Direct substitution results in $\frac{0}{0}$. We factor the numerator:
$$\lim_{x \to -3} \frac{(x+3)(x-2)}{x+3} = \lim_{x \to -3} (x-2) = -3 - 2 = -5$$
2. **Solution:** Divide the numerator and denominator by the highest power of $x$, which is
$x^3$:
$$\lim_{x \to \infty} \frac{\frac{5x^3}{x^3} - \frac{2x^2}{x^3} + \frac{1}{x^3}}{\frac{10x^3}{x^3}
+ \frac{3x}{x^3} - \frac{7}{x^3}} = \lim_{x \to \infty} \frac{5 - \frac{2}{x} + \frac{1}{x^3}}{10 + \
frac{3}{x^2} - \frac{7}{x^3}} = \frac{5-0+0}{10+0-0} = \frac{5}{10} = \frac{1}{2}$$
3. **Solution:** For $f(x)$ to be continuous at $x=2$, the left-hand limit must equal the right-
hand limit, and this must equal the function's value. We set the two pieces equal at $x=2$:
$$c(2)^2 + 2(2) = (2)^3 - c(2)$$
$$4c + 4 = 8 - 2c$$
$$6c = 4$$
$$c = \frac{4}{6} = \frac{2}{3}$$
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#### **Chapter 2 Solutions**
1. **Solution:**
$$f'(x) = \lim_{h \to 0} \frac{((x+h)^2 - 4(x+h)) - (x^2 - 4x)}{h}$$
$$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 4x - 4h - x^2 + 4x}{h}$$
$$= \lim_{h \to 0} \frac{2xh + h^2 - 4h}{h} = \lim_{h \to 0} \frac{h(2x + h - 4)}{h}$$
$$= \lim_{h \to 0} (2x + h - 4) = 2x - 4$$
2. **Solution:** Using the Product Rule $(fg)' = f'g + fg'$ and the Chain Rule:
Let $f(x) = e^{x^2}$ and $g(x) = \sin(3x)$.
Then $f'(x) = e^{x^2} \cdot 2x$ and $g'(x) = \cos(3x) \cdot 3$.
$$\frac{dy}{dx} = (2x e^{x^2})(\sin(3x)) + (e^{x^2})(3\cos(3x))$$
3. **Solution:** Differentiate both sides with respect to $x$:
$$\frac{d}{dx}(y^2) + \frac{d}{dx}(x^2y) = \frac{d}{dx}(5x)$$
$$2y \frac{dy}{dx} + (2xy + x^2 \frac{dy}{dx}) = 5$$
Group terms with $\frac{dy}{dx}$:
$$\frac{dy}{dx}(2y + x^2) = 5 - 2xy$$
$$\frac{dy}{dx} = \frac{5 - 2xy}{2y + x^2}$$
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#### **Chapter 3 Solutions**
1. **Solution:**
* $f'(x) = 6x^2 + 6x - 12 = 6(x^2+x-2) = 6(x+2)(x-1)$. Critical numbers are $x=-2, x=1$.
* Increasing ($f'(x)>0$): $(-\infty, -2)$ and $(1, \infty)$.
* Decreasing ($f'(x)<0$): $(-2, 1)$.
* Local max at $x=-2$, $f(-2)=21$. Local min at $x=1$, $f(1)=-6$.
2. **Solution:**
* $g'(x) = 4x^3 - 12x$.
* $g''(x) = 12x^2 - 12 = 12(x^2-1) = 12(x-1)(x+1)$. Zeros are $x=-1, x=1$.
* Concave upward ($g''(x)>0$): $(-\infty, -1)$ and $(1, \infty)$.
* Concave downward ($g''(x)<0$): $(-1, 1)$.
* Inflection points at $x=-1$ and $x=1$.
3. **Solution:** Let length be $x$ and width be $y$. Area $A = xy = 1,500,000$. The amount of
fence is $F = 2x + 3y$.
From area, $y = 1,500,000/x$. Substitute into $F$:
$F(x) = 2x + 3(\frac{1,500,000}{x}) = 2x + 4,500,000x^{-1}$.
$F'(x) = 2 - 4,500,000x^{-2}$. Set $F'(x)=0$:
$2 = \frac{4,500,000}{x^2} \implies x^2 = 2,250,000 \implies x = 1500$ ft.
Then $y = 1,500,000 / 1500 = 1000$ ft.
Dimensions are 1500 ft by 1000 ft.
4. **Solution:** Given $r=5$ (constant), $\frac{dV}{dt} = 3$ m³/min.
$V = \pi r^2 h = \pi (5)^2 h = 25\pi h$.
Differentiate with respect to $t$:
$\frac{dV}{dt} = 25\pi \frac{dh}{dt}$.
$3 = 25\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{3}{25\pi}$ m/min.
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#### **Chapter 4 Solutions**
1. **Solution:** The antiderivative of $\cos(x) + 3$ is $\sin(x) + 3x$.
$$\int_0^{\pi/2} (\cos(x) + 3) \,dx = [\sin(x) + 3x]_0^{\pi/2}$$
$$= (\sin(\pi/2) + 3(\pi/2)) - (\sin(0) + 3(0))$$
$$= (1 + \frac{3\pi}{2}) - (0 + 0) = 1 + \frac{3\pi}{2}$$
2. **Solution:** Let $u = x^2+1$. Then $du = 2x \,dx$, so $\frac{1}{2}du = x \,dx$.
$$\int \frac{\frac{1}{2}du}{u^2} = \frac{1}{2} \int u^{-2} \,du = \frac{1}{2} \frac{u^{-1}}{-1} + C$$
$$= -\frac{1}{2u} + C = -\frac{1}{2(x^2+1)} + C$$
3. **Solution:** First, find intersection points: $x^2 = 2x - x^2 \implies 2x^2 - 2x = 0 \implies
2x(x-1)=0$. Intersections are at $x=0, x=1$.
In the interval $[0, 1]$, the curve $y=2x-x^2$ is above $y=x^2$.
Area $A = \int_0^1 [(2x-x^2) - (x^2)] \,dx = \int_0^1 (2x - 2x^2) \,dx$.
$$A = [x^2 - \frac{2}{3}x^3]_0^1 = (1^2 - \frac{2}{3}(1)^3) - (0) = 1 - \frac{2}{3} = \frac{1}{3}$$
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