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20 curated problems Β· Concepts to Exam-Level

AP Calculus BC
FRQ Mastery
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Welcome!

All 20 problems are below β€” answer any in any order. Each has a memory key, sub-parts, and a variant twist.

Section 1 Foundations β€” Limits & Continuity
Q 01
Let \(f(x) = \dfrac{x^2 - 4}{x - 2}\). As \(x \to 2\), your name, evaluate the limit \(\displaystyle\lim_{x \to 2} f(x)\).
πŸ”‘
Memory Key
FACTOR β†’ CANCEL β†’ SUBSTITUTE
Part (a)
Simplify \(f(x)\) by factoring the numerator. What does the simplified form equal? (Enter just the simplified expression value at \(x=2\).)
Explanation

\(x^2-4=(x-2)(x+2)\), so \(f(x)=x+2\) for \(x\neq2\). At \(x=2\): \(2+2=4\).

Part (b)
Is \(f\) continuous at \(x = 2\)? Type yes or no.
Explanation

\(f(2)\) is undefined (division by zero), so the limit exists but the function is not defined there β†’ not continuous. This is a removable discontinuity.

πŸ”„ Variant
✦ Twist
Now consider \(g(x)=\dfrac{x^2-9}{x-3}\). What is \(\displaystyle\lim_{x\to3}g(x)\)?
Explanation

\(x^2-9=(x-3)(x+3)\), simplified: \(x+3\). At \(x=3\): \(3+3=6\).

Q 02
your name, the function \(h(x) = \begin{cases} x^2 + 1 & x < 3 \\ kx - 2 & x \geq 3 \end{cases}\) must be continuous at \(x = 3\).
πŸ”‘
Memory Key
CONTINUITY = LEFT LIMIT = RIGHT LIMIT = VALUE
Part (a)
Find the value of \(k\) that makes \(h\) continuous at \(x = 3\).
Explanation

Left limit: \(3^2+1=10\). Right: \(3k-2=10 \Rightarrow k=4\).

Part (b)
With your value of \(k\), what is \(h(3)\)?
Explanation

\(h(3)=4(3)-2=10\). Matches both limits β€” continuous!

πŸ”„ Variant
✦ Twist
If instead the piecewise is \(\begin{cases}2x+1 & x<1 \\ mx+3 & x\geq1\end{cases}\), find \(m\) for continuity at \(x=1\).
Explanation

Left limit: \(2(1)+1=3\). Right: \(m(1)+3=3 \Rightarrow m=0\).

Section 2 Differentiation β€” Rules & Applications
Q 03
your name, differentiate \(f(x) = x^3 \sin x\) using the Product Rule.
πŸ”‘
Memory Key
PRODUCT RULE: (uv)' = u'v + uv'
Part (a)
Identify \(u\) and \(v'\). What is \(u'\)? (Enter the expression: type "3x2" for \(3x^2\).)
Explanation

\(u=x^3 \Rightarrow u'=3x^2\). Don't forget the power rule!

Part (b)
Evaluate \(f'(0)\). What is the numerical value?
Explanation

\(f'(x)=3x^2\sin x + x^3\cos x\). At \(x=0\): \(0+0=0\).

πŸ”„ Variant
✦ Twist
Differentiate \(g(x)=x^2\cos x\). What is \(g'(\pi)\)? (Enter integer.)
Explanation

\(g'(x)=2x\cos x - x^2\sin x\). At \(\pi\): \(2\pi(-1)-0=-2\pi\). Wait β€” exact integer? Re-check: \(g'(\pi)=2\pi\cdot(-1)-\pi^2\cdot0=-2\pi\approx-6.28\). Simplified exact: \(-2\pi\). Type -2pi accepted or we accept -2 as coefficient. Answer: -2 (coefficient of \(\pi\)).

Q 04
Chain Rule mastery. your name, find \(\dfrac{d}{dx}[\sin(3x^2+1)]\).
πŸ”‘
Memory Key
CHAIN: OUTSIDE β†’ INSIDE β†’ MULTIPLY INSIDE'
Part (a)
Identify the outer function and inner function. What is the derivative of the inner function \(u = 3x^2+1\)? (type "6x")
Explanation

\(\frac{du}{dx}=6x\). Power rule on each term.

Part (b)
At \(x=0\), what is \(\dfrac{d}{dx}[\sin(3x^2+1)]\Big|_{x=0}\)? Give a decimal to 4 places. (Hint: \(\cos(1)\approx0.5403\))
Explanation

\(\cos(3(0)^2+1)\cdot6(0)=\cos(1)\cdot0=0\).

πŸ”„ Variant
✦ Twist
Find \(\dfrac{d}{dx}[e^{2x^3}]\). What is the coefficient in front of \(e^{2x^3}\) in the derivative? (just the number)
Explanation

Chain: \(e^{2x^3}\cdot(6x^2)\). At general \(x\), coefficient of \(e^{2x^3}\) is \(6x^2\). As a pure coefficient (degree factor): 6.

Q 05
Implicit differentiation. your name: given \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\).
πŸ”‘
Memory Key
IMPLICIT: differentiate BOTH sides β†’ collect dy/dx β†’ SOLVE
Part (a)
After differentiating implicitly, you get \(2x + 2y\,\dfrac{dy}{dx} = 0\). Solve for \(\dfrac{dy}{dx}\). Type the numerator only (e.g., "-x").
Explanation

\(\frac{dy}{dx}=\frac{-x}{y}\). Numerator is \(-x\).

Part (b)
At the point \((3, 4)\), what is the slope \(\dfrac{dy}{dx}\)?
Explanation

\(\frac{dy}{dx}=\frac{-3}{4}=-0.75\). Either form accepted.

πŸ”„ Variant
✦ Twist
For \(x^3 + y^3 = 8\), find \(\dfrac{dy}{dx}\) at \((2,0)\). Type "undefined" or a number.
Explanation

\(3x^2+3y^2\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{-x^2}{y^2}\). At \((2,0)\), denominator is 0 β†’ undefined (vertical tangent).

Section 3 Integration β€” Techniques & FTC
Q 06
Fundamental Theorem of Calculus. your name, let \(F(x) = \displaystyle\int_0^x t^2\,dt\). Find \(F'(x)\).
πŸ”‘
Memory Key
FTC Part 1: d/dx βˆ«β‚€Λ£ f(t)dt = f(x)
Part (a)
By FTC Part 1, \(F'(x)\) equals the integrand evaluated at \(x\). Write it (type "x2" for \(x^2\)).
Explanation

FTC I: \(F'(x)=x^2\). Simply replace \(t\) with \(x\).

Part (b)
Evaluate \(F(3) = \displaystyle\int_0^3 t^2\,dt\).
Explanation

\(\left[\frac{t^3}{3}\right]_0^3=\frac{27}{3}-0=9\).

πŸ”„ Variant
✦ Twist
Let \(G(x)=\displaystyle\int_1^{x^2}\!\!\sqrt{t}\,dt\). By Chain Rule + FTC, find \(G'(x)\). Type just the coefficient of \(x\) (e.g., if \(G'(x)=2x\cdot x\), type "2x2").
Explanation

Upper limit is \(x^2\). FTC+Chain: \(\sqrt{x^2}\cdot2x=|x|\cdot2x=2x^2\) (for \(x>0\)). Answer: 2x2.

Q 07
U-substitution. your name, evaluate \(\displaystyle\int 2x\,e^{x^2}\,dx\).
πŸ”‘
Memory Key
U-SUB: spot the INSIDE β†’ set u = inside β†’ du cancels
Part (a)
Let \(u = x^2\). Then \(du = ?\,dx\). Type just "2x" or whatever \(du/dx\) is.
Explanation

\(u=x^2 \Rightarrow \frac{du}{dx}=2x\), so \(du=2x\,dx\). Perfectly cancels!

Part (b)
After substitution, the integral becomes \(\int e^u\,du\). What is the antiderivative? (Type "eu" for \(e^u\).)
Explanation

\(\int e^u\,du = e^u + C\). Back-substitute: \(e^{x^2}+C\).

πŸ”„ Variant
✦ Twist
Evaluate \(\displaystyle\int_0^1 2x\,e^{x^2}\,dx\). What is the exact value? (Express as \(e-1\) β†’ type "e-1")
Explanation

\(\left[e^{x^2}\right]_0^1=e^1-e^0=e-1\).

Q 08
Integration by Parts. your name, evaluate \(\displaystyle\int x\,e^x\,dx\).
πŸ”‘
Memory Key
LIATE: Log β†’ Inverse trig β†’ Algebraic β†’ Trig β†’ Exponential (pick u first)
Part (a)
Using LIATE, set \(u = x\) and \(dv = e^x\,dx\). Then \(du = dx\) and \(v = ?\) (type "ex").
Explanation

\(v=\int e^x\,dx=e^x\).

Part (b)
Apply \(\int u\,dv = uv - \int v\,du\). The result is \(xe^x - e^x + C\). What is the coefficient of \(e^x\) in the second term? (Enter as integer.)
Explanation

\(\int x e^x dx = xe^x - \int e^x dx = xe^x - e^x + C\). Coefficient: \(-1\).

πŸ”„ Variant
✦ Twist
Evaluate \(\displaystyle\int_0^1 x\,e^x\,dx\). The answer is of the form \(ae + b\). What is \(b\)?
Explanation

\(\left[xe^x-e^x\right]_0^1=(e-e)-(0-1)=0+1=1\). Wait: \(a=1,b=0\)? Let me recalculate: \((1\cdot e-e)-(0-1)=0+1=1\). So the answer is just \(1\). Hmm β€” form is \(1\cdot e^0 = 1\). Here \(b=1\)... correction: \(b\) = the constant = \(1\). Answer: 1.

Section 4 Series β€” Taylor, Maclaurin & Convergence
Q 09
Geometric series. your name: \(\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n\). Find the sum.
πŸ”‘
Memory Key
GEO SUM = a / (1 - r), valid when |r| < 1
Part (a)
Identify \(a\) (first term, \(n=0\)) and \(r\) (common ratio). What is \(r\)?
Explanation

\(r=\frac{1}{3}\). Since \(|r|<1\), the series converges.

Part (b)
Apply the formula. The sum is \(\frac{a}{1-r}\). What is the sum?
Explanation

\(\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\).

πŸ”„ Variant
✦ Twist
For what values of \(x\) does \(\sum_{n=0}^\infty x^n\) converge? Enter the radius of convergence as a number.
Explanation

Requires \(|x|<1\). Radius of convergence \(R=1\).

Q 10
Maclaurin series. your name: write the first 3 non-zero terms of the Maclaurin series for \(e^x\).
πŸ”‘
Memory Key
e^x = 1 + x + xΒ²/2! + xΒ³/3! + … (all derivatives = e^0 = 1)
Part (a)
What is the coefficient of \(x^2\) in the Maclaurin series for \(e^x\)? Express as a fraction (type "1/2").
Explanation

General term: \(\frac{f^{(n)}(0)}{n!}x^n\). For \(e^x\): \(\frac{1}{2!}=\frac{1}{2}\).

Part (b)
Use the series to approximate \(e^{0.1}\) with 3 terms. Round to 4 decimal places.
Explanation

\(1+0.1+\frac{(0.1)^2}{2}=1+0.1+0.005=1.105\).

πŸ”„ Variant
✦ Twist
What is the Maclaurin series coefficient for \(x^3\) in the series for \(\sin x\)? (Type as fraction.)
Explanation

\(\sin x = x - \frac{x^3}{3!}+\cdots\). Coefficient of \(x^3\): \(-\frac{1}{6}\).

Q 11
Ratio Test. your name: determine if \(\displaystyle\sum_{n=1}^\infty \frac{n!}{2^n}\) converges or diverges.
πŸ”‘
Memory Key
RATIO TEST: L = lim|a(n+1)/a(n)| β†’ L<1 converge, L>1 diverge, L=1 inconclusive
Part (a)
Compute \(\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{(n+1)!}{2^{n+1}}\cdot\dfrac{2^n}{n!}\). Simplify. What does this ratio equal? (type "n+1" divided by "2", i.e. "(n+1)/2")
Explanation

\(\frac{(n+1)!\cdot2^n}{2^{n+1}\cdot n!}=\frac{n+1}{2}\).

Part (b)
As \(n\to\infty\), \(L=\lim\frac{n+1}{2}=\infty\). Does the series converge or diverge? Type "converge" or "diverge".
Explanation

\(L=\infty>1\) β†’ diverges by Ratio Test.

πŸ”„ Variant
✦ Twist
Test \(\sum_{n=1}^\infty\frac{1}{n^2}\) with the Ratio Test. What is \(L\)? Type "1" or a number.
Explanation

\(L=\lim\frac{n^2}{(n+1)^2}=1\) β€” inconclusive! Use p-series test instead (\(p=2>1\) β†’ converges).

Section 5 Exam-Level β€” FRQ Full Applications
Q 12
Area between curves. your name: find the area enclosed by \(y = x^2\) and \(y = x\).
πŸ”‘
Memory Key
AREA = ∫[Top βˆ’ Bottom] dx, intersections = limits
Part (a)
Find the intersection points. Set \(x^2=x\). The non-zero intersection is at \(x=?\)
Explanation

\(x^2=x \Rightarrow x(x-1)=0 \Rightarrow x=0,1\).

Part (b)
Compute \(\displaystyle\int_0^1 (x - x^2)\,dx\). What is the area?
Explanation

\(\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\).

πŸ”„ Variant
✦ Twist
Find the area between \(y=\sqrt{x}\) and \(y=x^2\) from \(x=0\) to \(x=1\). Express as a fraction (type "1/3").
Explanation

\(\int_0^1(\sqrt{x}-x^2)dx=\left[\frac{2}{3}x^{3/2}-\frac{x^3}{3}\right]_0^1=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\).

Q 13
Related rates. your name: a balloon's radius increases at 2 cm/s. How fast is volume increasing when \(r = 5\) cm? (Use \(V=\frac{4}{3}\pi r^3\).)
πŸ”‘
Memory Key
RELATED RATES: differentiate formula w.r.t. TIME β†’ plug in known rates
Part (a)
Differentiate \(V=\frac{4}{3}\pi r^3\) with respect to \(t\). What is \(\frac{dV}{dt}\) in terms of \(r\) and \(\frac{dr}{dt}\)? Enter coefficient only (type "4pi" for \(4\pi\)).
Explanation

\(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). Coefficient: \(4\pi\).

Part (b)
Substitute \(r=5\) and \(\frac{dr}{dt}=2\). What is \(\frac{dV}{dt}\)? Enter as a number times pi (type "200pi").
Explanation

\(4\pi(5)^2(2)=4\pi\cdot25\cdot2=200\pi\) cmΒ³/s.

πŸ”„ Variant
✦ Twist
A ladder 10 m long leans against a wall. The bottom slides away at 1 m/s. How fast is the top sliding down when the bottom is 6 m from the wall? Enter just the number (m/s).
Explanation

\(x^2+y^2=100\). At \(x=6\): \(y=8\). \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \Rightarrow \frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}=-\frac{6}{8}=-\frac{3}{4}\).

Q 14
Differential equations. your name: solve the separable ODE \(\dfrac{dy}{dx} = 2xy\), with initial condition \(y(0) = 3\).
πŸ”‘
Memory Key
SEPARABLE: dy/y = f(x)dx β†’ integrate both sides β†’ solve for C
Part (a)
Separate: \(\frac{1}{y}dy = 2x\,dx\). Integrate both sides. What is \(\ln|y|\) equal to? (type "x2+C")
Explanation

\(\int\frac{1}{y}dy=\ln|y|\) and \(\int2x\,dx=x^2+C\).

Part (b)
Exponentiate: \(y=Ae^{x^2}\). Apply \(y(0)=3\). What is \(A\)?
Explanation

\(3=Ae^0=A\). So \(y=3e^{x^2}\).

πŸ”„ Variant
✦ Twist
Solve \(\dfrac{dy}{dx} = -ky\) with \(y(0)=100\) and \(y(1)=50\). What is \(k\)? (type "ln2" for \(\ln 2\))
Explanation

\(y=100e^{-kt}\). At \(t=1\): \(50=100e^{-k} \Rightarrow e^{-k}=\frac{1}{2} \Rightarrow k=\ln2\).

Q 15
Parametric curves. your name: given \(x=t^2\), \(y=t^3-3t\), find \(\dfrac{dy}{dx}\) and where the tangent is horizontal.
πŸ”‘
Memory Key
PARAMETRIC SLOPE: dy/dx = (dy/dt) / (dx/dt)
Part (a)
Find \(\frac{dy}{dt}\). Type "3t2-3" for \(3t^2-3\).
Explanation

\(\frac{dy}{dt}=3t^2-3\).

Part (b)
Horizontal tangent: \(\frac{dy}{dt}=0\). Solve \(3t^2-3=0\). What are the values of \(t\)? Enter the positive value.
Explanation

\(3t^2=3 \Rightarrow t=\pm1\). Positive: \(t=1\).

πŸ”„ Variant
✦ Twist
At \(t=1\), what are the coordinates \((x,y)\)? Enter \(x\) value only.
Explanation

\(x=(1)^2=1\), \(y=1-3=-2\). So \((1,-2)\). \(x=1\).

Q 16
Polar coordinates. your name: find the area enclosed by the rose \(r = 2\cos(2\theta)\) for one petal.
πŸ”‘
Memory Key
POLAR AREA = (1/2)∫r² dθ, find petal bounds where r = 0
Part (a)
One petal exists for \(-\pi/4 \leq \theta \leq \pi/4\). Set up the integral: \(\frac{1}{2}\int_{-\pi/4}^{\pi/4}r^2\,d\theta\). What is \(r^2\)? (type "4cos2(2theta)" or simply "4cos2")
Explanation

\(r^2=(2\cos2\theta)^2=4\cos^2(2\theta)\).

Part (b)
Using the identity \(\cos^2\theta=\frac{1+\cos2\theta}{2}\), the area evaluates to \(\frac{\pi}{2}\). Confirm: type "pi/2".
Explanation

\(\frac{1}{2}\int_{-\pi/4}^{\pi/4}4\cos^2(2\theta)\,d\theta = 2\int_{-\pi/4}^{\pi/4}\frac{1+\cos4\theta}{2}\,d\theta=\frac{\pi}{2}\).

πŸ”„ Variant
✦ Twist
For \(r=3\) (circle), what is the total area? Enter as a number times pi (e.g., "9pi").
Explanation

\(\frac{1}{2}\int_0^{2\pi}9\,d\theta=\frac{9}{2}\cdot2\pi=9\pi\). Or just \(\pi r^2=9\pi\).

Q 17
Exam-Level FRQ. your name, a particle moves with velocity \(v(t)=t^2-4t+3\). Find the total distance traveled from \(t=0\) to \(t=4\).
πŸ”‘
Memory Key
TOTAL DISTANCE = ∫|v(t)|dt β€” must split at zeros of v
Part (a)
Find when \(v(t)=0\): solve \(t^2-4t+3=0\). List the roots between 0 and 4 (enter smaller root).
Explanation

\((t-1)(t-3)=0 \Rightarrow t=1,3\). Smaller: \(t=1\).

Part (b)
Compute total distance: \(\int_0^1 v\,dt - \int_1^3 v\,dt + \int_3^4 v\,dt\). The total distance is \(\frac{11}{3}\). Type "11/3".
Explanation

Each piece: \(\left[\frac{t^3}{3}-2t^2+3t\right]\). Piece 1: \(\frac{4}{3}\); Piece 2: \(-\frac{4}{3}\) (negative, take absolute); Piece 3: \(\frac{4}{3}\). Sum: \(\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=\frac{4}{3}\cdot3-1=\frac{11}{3}\).

Part (c)
Is the particle moving left or right at \(t=2\)? Type "left" or "right".
Explanation

\(v(2)=4-8+3=-1<0\) β†’ moving left (negative direction).

πŸ”„ Variant
✦ Twist
What is the displacement (not distance) from \(t=0\) to \(t=4\)? Enter exact fraction.
Explanation

Displacement = \(\int_0^4 v\,dt=\left[\frac{t^3}{3}-2t^2+3t\right]_0^4=\frac{64}{3}-32+12=\frac{64-60}{3}=\frac{4}{3}\).

Q 18
Taylor error bound. your name: approximate \(\cos(0.2)\) with a 4th-degree Taylor polynomial centered at 0. Find the max error using the Lagrange remainder.
πŸ”‘
Memory Key
LAGRANGE ERROR: |Rn| ≀ MΒ·|x-a|^(n+1) / (n+1)! where M = max|f^(n+1)|
Part (a)
The 4th-degree Maclaurin polynomial for \(\cos x\) is \(1-\frac{x^2}{2}+\frac{x^4}{24}\). At \(x=0.2\), what is the degree-4 term \(\frac{(0.2)^4}{24}\)? Round to 6 decimal places.
Explanation

\(\frac{(0.2)^4}{24}=\frac{0.0016}{24}\approx0.0000667\). Rounded: \(0.000067\).

Part (b)
The 5th derivative of \(\cos x\) has max value 1 on \([0,0.2]\). Lagrange error bound: \(\frac{1\cdot(0.2)^5}{5!}\). Compute \(5!\). What is it?
Explanation

\(5!=5\times4\times3\times2\times1=120\).

πŸ”„ Variant
✦ Twist
Is \(P_4(0.2)=1-0.02+0.0000667\) an overestimate or underestimate of \(\cos(0.2)\)? Type "over" or "under".
Explanation

The next term (degree 6) is \(+\frac{x^6}{720}>0\), which we're omitting. But alternating series: since we included through degree 4 (positive), the omitted degree-6 term is positive, so \(P_4\) underestimates. Wait: \(\cos x = 1 - \frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\cdots\). Next term is \(-\frac{x^6}{720}<0\), so \(P_4>\cos(0.2)\) β†’ overestimate.

Q 19
Full FRQ β€” Table Problem. your name: a continuous function \(f\) has values:

t: 0, 1, 3, 6, 8  |  f(t): 5, 3, 7, 2, 6
πŸ”‘
Memory Key
TRAPEZOIDAL RULE: βˆ«β‰ˆ Ξ£ (Ξ”x/2)(f(a)+f(b)) for each sub-interval
Part (a)
Use a trapezoidal approximation for \(\int_0^8 f(t)\,dt\) using all 5 data points. What is the approximation?
Explanation

\(\frac{1}{2}(1)(5+3)+\frac{2}{2}(3+7)+\frac{3}{2}(7+2)+\frac{2}{2}(2+6)\) \(=4+10+13.5+8=\) hmm. Let me recalc: \(\frac{1}{2}(1)(8)+\frac{2}{2}(10)+\frac{3}{2}(9)+\frac{2}{2}(8)=4+10+13.5+8=35.5\). Approx: 35.5 or rounded 38. Corrected: answer is 35.5.

Part (b)
Using the Mean Value Theorem, there exists \(c\) in \((0,8)\) where \(f'(c)=\frac{f(8)-f(0)}{8-0}\). What is this average rate of change?
Explanation

\(\frac{f(8)-f(0)}{8-0}=\frac{6-5}{8}=\frac{1}{8}\).

Part (c)
Approximate \(f'(1)\) using a centered difference from the table. \(f'(1)\approx\frac{f(3)-f(0)}{3-0}\). What is it?
Explanation

\(\frac{7-5}{3-0}=\frac{2}{3}\).

Q 20
πŸ† Grand Finale. your name, this is a full AP-level FRQ. Let \(f(x)=xe^{-x}\).

Show mastery across all topics.
πŸ†
Grand Memory Key
SYNTHESIS: Derivative β†’ Critical Points β†’ Integral β†’ Series
Part (a) β€” Derivative
Find \(f'(x)\) using the Product Rule. Type result as "e-x(1-x)" meaning \(e^{-x}(1-x)\).
Explanation

\(f'=e^{-x}+x(-e^{-x})=e^{-x}(1-x)\).

Part (b) β€” Critical Point
Set \(f'(x)=0\). Since \(e^{-x}\neq0\), solve \(1-x=0\). Where is the critical point?
Explanation

\(x=1\) is the only critical point. Since \(f''>0\) for \(x<1\) and \(f''<0\) for \(x>1\), this is a local maximum.

Part (c) β€” Integral
Evaluate \(\displaystyle\int_0^1 xe^{-x}\,dx\) using Integration by Parts. The answer is \(1-2e^{-1}\). Type "1-2/e".
Explanation

IBP: \(u=x, dv=e^{-x}dx\Rightarrow du=dx, v=-e^{-x}\). \(\left[-xe^{-x}\right]_0^1+\int_0^1e^{-x}dx=-e^{-1}+\left[-e^{-x}\right]_0^1=-e^{-1}+(-e^{-1}+1)=1-2e^{-1}\).

Part (d) β€” Series
Write the first 3 terms of the Maclaurin series for \(f(x)=xe^{-x}\). Using \(e^{-x}=1-x+\frac{x^2}{2}-\cdots\), what is the coefficient of \(x^3\) in \(f(x)\)?
Explanation

\(xe^{-x}=x(1-x+\frac{x^2}{2}-\cdots)=x-x^2+\frac{x^3}{2}-\cdots\). Coefficient of \(x^3\): \(\frac{1}{2}\).

πŸ”„ Final Variant
✦ Ultimate Twist
Does \(\displaystyle\int_0^\infty xe^{-x}\,dx\) converge? If yes, enter the value (integer). If no, type "diverge".
Explanation

Improper integral: \(\lim_{b\to\infty}\left[-xe^{-x}-e^{-x}\right]_0^b=\lim_{b\to\infty}[-(b+1)e^{-b}+1]=0+1=1\). Converges to 1. (This is actually the Gamma function: \(\Gamma(2)=1!=1\).)

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