How to Use This Practice Final
This page is a full-length practice final for a first course in calculus. It contains thirty problems, three from each of the ten units of the course, and every problem is followed by a complete worked solution. The problems are ordered the way the course is ordered, so you can work through them in sequence or jump to the unit you want to review.
For the most realistic practice, set a timer for three hours, close the solutions, and work as many problems as you can on paper. A typical final exam contains eight to twelve problems of this size, so you are not expected to finish all thirty in one sitting. A better plan is to do one unit at a time: attempt the three problems, check your work against the solutions, and then read the tutorials listed at the end of the unit for any method that felt shaky.
Unit 1: Function Models and Computational Foundations
Problem 1 · Exponential Models
An account grows with continuous compounding at \(4\%\) per year, so its balance is \(A(t)=A_0e^{0.04t}\). How long does it take to double?
Solution. Set \(A(t)=2A_0\). The initial amount cancels, leaving \(e^{0.04t}=2\). Taking natural logarithms,
Answer: \(t\approx17.33\) years.
Problem 2 · Trigonometric Equations
Solve \(2\sin^2x-\sin x-1=0\) for \(0\le x<2\pi\).
Solution. This is a quadratic in \(s=\sin x\): \(2s^2-s-1=(2s+1)(s-1)=0\). So \(\sin x=1\) or \(\sin x=-\frac12\). On \([0,2\pi)\), \(\sin x=1\) gives \(x=\frac\pi2\). The value \(-\frac12\) occurs in the third and fourth quadrants, at reference angle \(\frac\pi6\), giving \(x=\pi+\frac\pi6\) and \(x=2\pi-\frac\pi6\).
Answer: \(x=\dfrac\pi2,\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}\).
Problem 3 · Composition and Domain
Let \(f(x)=\sqrt{x-1}\) and \(g(x)=\dfrac{1}{x-3}\). Find the domain of \(f\circ g\).
Solution. We have \((f\circ g)(x)=\sqrt{\frac{1}{x-3}-1}=\sqrt{\frac{4-x}{x-3}}\). Two conditions apply: \(x\ne3\) (for \(g\)), and the radicand must satisfy \(\frac{4-x}{x-3}\ge0\). A sign chart for the quotient: it is negative for \(x<3\), positive for \(3<x<4\), zero at \(x=4\), and negative for \(x>4\). So the radicand is nonnegative exactly when \(3<x\le4\).
Answer: The domain is \((3,4]\).
Review these tutorials: Relations and the Function Rule; Polynomial Standard Form and Degree; Integer Exponents and Repeated Multiplication; Radians as Angle Measure.
Unit 2: Limits and Continuity
Problem 4 · An Algebraic Limit
Find \(\lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x}\).
Solution. Direct substitution gives \(\frac00\). Multiply the numerator and denominator by the conjugate \(\sqrt{1+x}+1\):
As \(x\to0\) this tends to \(\frac{1}{1+1}\).
Answer: \(\dfrac12\).
Problem 5 · A Limit at Infinity
Find \(\lim_{x\to\infty}\dfrac{3x^2+x}{\sqrt{x^4+1}}\).
Solution. Divide the numerator and the denominator by \(x^2\), the highest power that matters. For \(x>0\), \(\sqrt{x^4+1}/x^2=\sqrt{1+x^{-4}}\), so
Answer: \(3\).
Problem 6 · Continuity and the IVT
(a) Find \(k\) so that \(f(x)=\begin{cases}\frac{x^2-16}{x-4},&x\ne4\\ k,&x=4\end{cases}\) is continuous at \(x=4\). (b) Show that \(x^3+x-1=0\) has a solution in \((0,1)\).
Solution. (a) For \(x\ne4\), \(\frac{x^2-16}{x-4}=x+4\), so \(\lim_{x\to4}f(x)=8\). Continuity requires \(f(4)=k=8\). (b) The function \(h(x)=x^3+x-1\) is a polynomial, hence continuous everywhere. We have \(h(0)=-1<0\) and \(h(1)=1>0\). By the Intermediate Value Theorem, \(h\) takes the value \(0\) at some point in \((0,1)\).
Answer: (a) \(k=8\). (b) A root exists in \((0,1)\); it is approximately \(0.682\).
Review these tutorials: Local Behavior Near an Input; Sum and Difference Laws; One-Sided Limit Notation; Three Conditions for Continuity at a Point.
Unit 3: Derivative Foundations
Problem 7 · The Limit Definition
Use the limit definition to find \(f'(2)\) for \(f(x)=\dfrac1x\).
Solution. By definition, \(f'(2)=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}\). Compute the difference quotient with a common denominator:
Letting \(h\to0\) gives \(\frac{-1}{4}\).
Answer: \(f'(2)=-\dfrac14\).
Problem 8 · Tangent Lines
Find the equation of the tangent line to \(y=x^3-2x\) at \(x=1\).
Solution. At \(x=1\), \(y=1-2=-1\), so the point is \((1,-1)\). The derivative is \(y'=3x^2-2\), so the slope is \(3-2=1\). Point-slope form gives \(y+1=1\cdot(x-1)\).
Answer: \(y=x-2\).
Problem 9 · Power Rule
Differentiate \(f(x)=3x^4-2\sqrt x+\dfrac5x\).
Solution. Rewrite with exponents: \(f(x)=3x^4-2x^{1/2}+5x^{-1}\). The power rule applies term by term:
Answer: \(f'(x)=12x^3-\dfrac{1}{\sqrt x}-\dfrac{5}{x^2}\).
Review these tutorials: Average Rate Of Change As A Secant Slope; Derivative as a Limit of Secant Slopes; Derivative Functions As Input-Dependent Rates; Constant Function Derivatives.
Unit 4: Differentiation Techniques
Problem 10 · Product and Chain Rules
Differentiate \(y=e^{2x}\cos3x\).
Solution. Use the product rule, with the chain rule on each factor: \(\frac{d}{dx}e^{2x}=2e^{2x}\) and \(\frac{d}{dx}\cos3x=-3\sin3x\). Then
Answer: \(y'=e^{2x}\,(2\cos3x-3\sin3x)\).
Problem 11 · Implicit Differentiation
Find \(\frac{dy}{dx}\) at the point \((2,1)\) on the curve \(x^2+xy+y^2=7\).
Solution. First check that \((2,1)\) lies on the curve: \(4+2+1=7\). Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\):
At \((2,1)\): \(-\frac{4+1}{2+2}\).
Answer: \(\dfrac{dy}{dx}=-\dfrac54\).
Problem 12 · Logarithmic Differentiation
Find \(\frac{d}{dx}\,x^x\) for \(x>0\), and evaluate it at \(x=2\).
Solution. Neither the power rule nor the exponential rule applies, since both the base and the exponent vary. Let \(y=x^x\); then \(\ln y=x\ln x\). Differentiating implicitly, \(\frac{y'}{y}=\ln x+x\cdot\frac1x=\ln x+1\). So
At \(x=2\), \(2^2(\ln2+1)=4(\ln2+1)\).
Answer: \(x^x(\ln x+1)\); at \(x=2\) it equals \(4(1+\ln2)\approx6.77\).
Review these tutorials: Deriving the Product Rule from First Principles; Deriving the Sine Derivative from First Principles; The Compound-Growth Limit Defining e; Implicit Relations And Local Function Branches.
Unit 5: Local Applications of Derivatives
Problem 13 · Linearization
Use the linearization of \(f(x)=\sqrt x\) at \(a=25\) to approximate \(\sqrt{26}\).
Solution. We have \(f(25)=5\) and \(f'(x)=\frac{1}{2\sqrt x}\), so \(f'(25)=\frac1{10}\). The linearization is \(L(x)=5+\frac1{10}(x-25)\). Then \(L(26)=5+0.1\).
Answer: \(\sqrt{26}\approx5.1\). The true value is \(5.0990\ldots\), so the estimate is slightly high, as expected for a concave function.
Problem 14 · Related Rates
A \(13\)-foot ladder leans against a vertical wall. The bottom slides away from the wall at \(2\) ft/s. How fast is the top sliding down when the bottom is \(5\) feet from the wall?
Solution. Let \(x\) be the distance from the wall to the bottom and \(y\) the height of the top, so \(x^2+y^2=169\). When \(x=5\), \(y=12\). Differentiate with respect to time:
Answer: \(\dfrac{dy}{dt}=-\dfrac{5}{6}\) ft/s, so the top is sliding down at \(\frac56\approx0.83\) ft/s.
Problem 15 · L'Hôpital's Rule
Find \(\lim_{x\to0}\dfrac{e^x-1-x}{x^2}\).
Solution. Direct substitution gives \(\frac00\). Apply L'Hôpital's rule: the quotient of derivatives is \(\frac{e^x-1}{2x}\), which is again \(\frac00\) at \(0\). Apply the rule once more:
Answer: \(\dfrac12\).
Review these tutorials: Tangent-Line Models Near a Base Point; Assigning Time Variables to Changing Quantities; Secant Slopes and Instantaneous Slopes; Recognizing Indeterminate Forms by Substitution.
Unit 6: Global Behavior and Optimization
Problem 16 · Critical Points
Find and classify the local extrema of \(f(x)=x^3-3x^2-9x+5\).
Solution. The derivative is \(f'(x)=3x^2-6x-9=3(x-3)(x+1)\), so the critical points are \(x=-1\) and \(x=3\). The sign of \(f'\) is positive for \(x<-1\), negative on \((-1,3)\), and positive for \(x>3\). So \(f\) increases then decreases at \(x=-1\) (a local maximum), and decreases then increases at \(x=3\) (a local minimum). The values are \(f(-1)=-1-3+9+5=10\) and \(f(3)=27-27-27+5=-22\).
Answer: Local maximum \(f(-1)=10\); local minimum \(f(3)=-22\).
Problem 17 · Concavity and Inflection
Find the inflection points of \(f(x)=x^4-4x^3\).
Solution. We have \(f'(x)=4x^3-12x^2\) and \(f''(x)=12x^2-24x=12x(x-2)\). Then \(f''=0\) at \(x=0\) and \(x=2\). The sign of \(f''\) is positive for \(x<0\), negative on \((0,2)\), and positive for \(x>2\). The concavity changes at both points, so both are inflection points. The \(y\)-values are \(f(0)=0\) and \(f(2)=16-32=-16\).
Answer: Inflection points at \((0,0)\) and \((2,-16)\).
Problem 18 · Optimization
An open box is made from a \(12\)-inch square sheet by cutting a square of side \(x\) from each corner and folding up the sides. Find the largest possible volume.
Solution. The base is \((12-2x)\) on a side and the height is \(x\), so \(V(x)=x(12-2x)^2\) on \(0\le x\le6\). Then
The critical points are \(x=6\) and \(x=2\). At the endpoints \(x=0\) and \(x=6\), \(V=0\). At \(x=2\), \(V=2\cdot8^2=128\), which is the maximum.
Answer: The largest volume is \(128\) cubic inches, with \(x=2\) inches.
Review these tutorials: Local Maximum and Minimum Definitions; Concavity Defined by Chord Slopes; Curve Analysis Data Tables; Translating Optimization Descriptions Into Constraints.
Unit 7: Accumulation and Definite Integrals
Problem 19 · Riemann Sums
Evaluate \(\lim_{n\to\infty}\sum_{i=1}^n\left(\dfrac in\right)^2\dfrac1n\) by recognizing it as a definite integral.
Solution. The sum is a right-endpoint Riemann sum for \(f(x)=x^2\) on \([0,1]\) with \(\Delta x=\frac1n\) and sample points \(x_i=\frac in\). Therefore the limit is the integral
Answer: \(\dfrac13\).
Problem 20 · Integrals from Geometry
Evaluate \(\int_{-3}^{3}\sqrt{9-x^2}\,dx\) without finding an antiderivative.
Solution. The graph of \(y=\sqrt{9-x^2}\) is the upper half of the circle \(x^2+y^2=9\), of radius \(3\). The integral is the area under this curve, which is a half-disk: \(\frac12\pi(3)^2\).
Answer: \(\dfrac{9\pi}{2}\).
Problem 21 · Properties of the Integral
Suppose \(\int_0^5f(x)\,dx=7\), \(\int_2^5f(x)\,dx=3\), and \(\int_0^2g(x)\,dx=-3\). Find \(\int_0^2\bigl(2f(x)-g(x)\bigr)\,dx\).
Solution. By additivity, \(\int_0^2f=\int_0^5f-\int_2^5f=7-3=4\). By linearity,
Answer: \(11\).
Review these tutorials: Accumulation as a Sum of Small Changes; Rectangular Approximations and the Definite Integral; Signed Integral Orientation; Variable-Limit Integrals Define Functions.
Unit 8: Antiderivatives and the Fundamental Theorem
Problem 22 · The FTC, Part 1
Find \(\dfrac{d}{dx}\displaystyle\int_1^{x^2}e^{t^2}\,dt\).
Solution. Let \(G(u)=\int_1^ue^{t^2}\,dt\). By the Fundamental Theorem, \(G'(u)=e^{u^2}\). The integral in the problem is \(G(x^2)\), so the chain rule gives
Answer: \(2x\,e^{x^4}\).
Problem 23 · Substitution
Evaluate \(\int_0^{\pi/2}\sin x\cos^2x\,dx\).
Solution. Let \(u=\cos x\), so \(du=-\sin x\,dx\). The limits change: \(x=0\) gives \(u=1\), and \(x=\frac\pi2\) gives \(u=0\). Then
Answer: \(\dfrac13\).
Problem 24 · Substitution with a Radical
Evaluate \(\int_0^1\dfrac{x}{\sqrt{1+x^2}}\,dx\).
Solution. Let \(u=1+x^2\), so \(du=2x\,dx\) and \(x\,dx=\frac12du\). The limits become \(u=1\) and \(u=2\):
Answer: \(\sqrt2-1\approx0.414\).
Review these tutorials: Antiderivatives And Reverse Differentiation; The Two Parts of the Fundamental Theorem; Substitution As The Reverse Chain Rule; Choosing an Evaluation Route for Mixed Integrals.
Unit 9: Applications of Definite Integrals
Problem 25 · Area Between Curves
Find the area of the region bounded by \(y=x^2\) and \(y=2x\).
Solution. Set \(x^2=2x\): the curves meet at \(x=0\) and \(x=2\). On \((0,2)\), the line \(y=2x\) lies above the parabola (at \(x=1\): \(2>1\)). So
Answer: \(\dfrac43\).
Problem 26 · Volume by Disks
The region under \(y=\sqrt x\) for \(0\le x\le4\) is rotated about the \(x\)-axis. Find the volume.
Solution. Each cross-section is a disk of radius \(\sqrt x\), with area \(\pi(\sqrt x)^2=\pi x\). Thus
Answer: \(8\pi\approx25.13\).
Problem 27 · Average Value
Find the average value of \(f(x)=\sin x\) on \([0,\pi]\).
Solution. The average value is \(\frac{1}{b-a}\int_a^bf\,dx\):
Answer: \(\dfrac2\pi\approx0.637\).
Review these tutorials: Deriving Area From Vertical Slices; Slicing Solids Into Cross-Sections; Definite Integrals as Totals of Variable Rates; From Variable Force to a Work Integral.
Unit 10: Differential Equations and Numerical Modeling
Problem 28 · Separable Equations
Solve \(\dfrac{dy}{dx}=xy\) with \(y(0)=2\).
Solution. Separate variables (for \(y\ne0\)): \(\frac{dy}{y}=x\,dx\). Integrate both sides: \(\ln|y|=\frac{x^2}{2}+C\), so \(y=Ae^{x^2/2}\). The initial condition \(y(0)=2\) gives \(A=2\).
Answer: \(y=2e^{x^2/2}\).
Problem 29 · Decay and Half-Life
A substance decays exponentially, and \(70\%\) of it remains after \(10\) years. Find its half-life.
Solution. Write \(A(t)=A_0e^{-kt}\). Then \(0.7=e^{-10k}\), so \(k=-\frac{\ln0.7}{10}\approx0.035667\). The half-life \(T\) satisfies \(e^{-kT}=\frac12\), that is,
Answer: \(T\approx19.43\) years.
Problem 30 · Euler's Method
Use Euler's method with step size \(h=0.1\) to approximate \(y(0.2)\) for \(y'=x+y\), \(y(0)=1\).
Solution. Euler's method updates \(y_{n+1}=y_n+h\,f(x_n,y_n)\) with \(f(x,y)=x+y\). Starting at \((x_0,y_0)=(0,1)\):
Answer: \(y(0.2)\approx1.22\). The exact solution is \(y=2e^x-x-1\), which gives \(1.2428\); the approximation is low because Euler's method underestimates a convex solution.
Review these tutorials: First-Order Differential Equation Notation; Constant Relative Rate as an Exponential Model; Constructing Euler Inputs from an Initial-Value Problem; Stock-Rate-Total Model Architecture.
Answer Key at a Glance
| Problem | Answer | Problem | Answer |
|---|---|---|---|
| 1 | \(\approx17.33\) years | 16 | max \(10\) at \(-1\); min \(-22\) at \(3\) |
| 2 | \(\frac\pi2,\frac{7\pi}6,\frac{11\pi}6\) | 17 | \(x=0\) and \(x=2\) |
| 3 | \((3,4]\) | 18 | \(128\) in\(^3\) |
| 4 | \(\frac12\) | 19 | \(\frac13\) |
| 5 | \(3\) | 20 | \(\frac{9\pi}2\) |
| 6 | \(k=8\); root in \((0,1)\) | 21 | \(11\) |
| 7 | \(-\frac14\) | 22 | \(2xe^{x^4}\) |
| 8 | \(y=x-2\) | 23 | \(\frac13\) |
| 9 | \(12x^3-x^{-1/2}-5x^{-2}\) | 24 | \(\sqrt2-1\) |
| 10 | \(e^{2x}(2\cos3x-3\sin3x)\) | 25 | \(\frac43\) |
| 11 | \(-\frac54\) | 26 | \(8\pi\) |
| 12 | \(x^x(\ln x+1)\); \(4(1+\ln2)\) | 27 | \(\frac2\pi\) |
| 13 | \(5.1\) | 28 | \(y=2e^{x^2/2}\) |
| 14 | \(-\frac56\) ft/s | 29 | \(\approx19.43\) years |
| 15 | \(\frac12\) | 30 | \(1.22\) |
What to Do After the Practice Final
Sort the problems you missed into three groups. The first group is errors of method, where you did not know which technique to use. For these, return to the review tutorials listed at the end of the unit and work a few additional examples before attempting the problem again. The second group is errors of execution, such as a lost sign, a forgotten chain-rule factor, or a wrong endpoint. These are best fixed by building checking habits: verify every derivative against a nearby value, estimate every integral's size before computing it, and substitute every endpoint twice. The third group is errors of recall, such as a forgotten derivative formula or trigonometric value. A one-page formula sheet that you rewrite from memory each day for a week closes this gap quickly.
Repeat the weakest unit a few days later with fresh problems of your own design, changing the numbers in the problems above. If you can solve the variations without looking at the solution, the technique has settled in. On a cumulative final, the skill that matters most is recognizing which tool a problem calls for: a limit, a derivative rule, an optimization setup, or an integral method.
