Final Exam Review

$\newenvironment {prompt}{}{} \newcommand {\bart }[0]{BART} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

In this section we prepare for the final exam.

Limits

$\lim _{x \to 2} \frac {x^2 - 4}{x^2 - 5x + 6} = \answer {-4}.$

$\lim _{x \to 3} \frac {x^2 -x -6}{x^2 - 9} = \answer {5/6}.$

$\lim _{x \to 4} \frac {\sqrt x - 2}{x^2 - 5x + 4} = \answer {1/12}.$

$\lim _{x \to 9} \frac {\sqrt x - 3}{x^2 - 8x - 9} = \answer {1/60}.$

$\lim _{x \to 2} \frac {\frac {1}{x} - \frac 12 }{x- 2} = \answer {-1/4}.$

$\lim _{x \to {-3}} \frac {\frac {1}{x} + \frac 13 }{x+ 3} = \answer {-1/9}.$

$\lim _{x \to \infty } \frac {x^2 - 5x + 4}{2x^2 + 4x + 7} = \answer {1/2}.$

$\lim _{x \to -\infty } \frac {3x^2 + x - 2}{x^3 -3x + 1} = \answer {0}.$

$\lim _{x \to 4^+} \frac {2x}{x-4} = \answer {\infty }.$

$\lim _{x \to 2^-} \frac {x+1}{x^2 - 4} = \answer {-\infty }.$

$\lim _{x \to 0} \frac {\sin (3x)}{x} = \answer {3}.$

$\lim _{x \to 0} \frac {e^{2x} -1}{x} = \answer {2}.$

$\lim _{x \to \infty } \frac {\ln (x)}{x} = \answer {0}.$

$\lim _{x \to \infty } \frac {e^x}{x^2} = \answer {\infty }.$

Derivatives

$\frac {d}{dx}\left (5x^4 - 7x^2 + 3x -3 \right ) = \answer {20x^3 - 14x + 3}.$

$\frac {d}{dx}\left (x^3 + \frac {1}{x^3} + \sqrt [3]x \right ) = \answer {3x^2 -3x^{-4} +(1/3)x^{-2/3}}.$

$\frac {d}{dx}\big [\tan (x) + \cot (x) \big ] = \answer {\sec ^2(x) - \csc ^2(x)}.$

$\frac {d}{dx}\big [\sec (x) + \csc (x) \big ] = \answer {\sec (x)\tan (x) - \csc (x)\cot (x)}.$

$\frac {d}{dx}\left (2^x + e^x \right ) = \answer {2^x \ln (2) + e^x}.$

$\frac {d}{dx}\big [\tan ^{-1}(x) + \sin ^{-1}(x) \big ] = \answer [given]{\frac {1}{1+x^2} + \frac {1}{(1-x^2)^{1/2}}}.$

$\frac {d}{dx}\left (x^2 \sin (x) + \sin (x^2) \right ) = \answer {x^2\cos (x) + 2x\sin (x) + 2x\cos (x^2)}.$

$\frac {d}{dx}(x^3 + 2x^2 - 5x - 4 )^6 = \answer {6(x^3 + 2x^2 - 5x - 4 )^5(3x^2 + 4x -5)}.$

$\frac {d}{dx}\left (x^3 \ln (x) \right ) = \answer {x^2 + 3x^2\ln (x)}.$

$\frac {d}{dx}\left (e^{3x} \tan (2x)\right ) = \answer {3e^{3x} \tan (2x) + 2e^{3x}\sec ^2(2x)}.$

$\frac {d}{dx}\left (\frac {2x+5}{3x - 4}\right ) = \answer {\frac {-23}{(3x-4)^2}}.$

$\frac {d}{dx}\left (\frac {x}{x^2 + 1}\right ) = \answer {\frac {1 - x^2}{(x^2 + 1)^2}}.$

Find the equation of the tangent line to the graph of $y=\sqrt {x}$ at $x = 16$ , and use it to approximate $\sqrt {18}$ .
The equation is $y = \answer {x/8 + 2}$ and $\sqrt {18} \approx \answer {4.25}$ .

Find the equation of the tangent line to the graph of $y=1/x$ at $x = 3$ .
The equation is $y = \answer {-x/9 +2/3}$ .

Integrals

$\int \left (x^2 - 5x + 3 \right ) \ dx = \answer {x^3/3 - 5x^2 /2 + 3x} + C.$

$\int \left (x^3 + \frac {1}{x^3} + \sqrt [3]x \right ) \ dx = \answer {x^4/4 +x^{-2}/{-2} + (3/4)x^{4/3}} + C.$

$\int \big [2\sin (x) - 3\cos (x) \big ] \ dx = \answer {-2\cos (x) -3\sin (x)} + C.$

$\int \big [4\sec ^2(x) + 5 \csc ^2(x) \big ] \ dx = \answer {4\tan (x) - 5 \cot (x)} + C.$

$\int x^2 \cos (x^3 + 1) \ dx = \answer {(1/3) \sin (x^3 +1)} + C.$

$\int \frac {\ln ^3(x)}{x} \ dx = \answer {\ln ^4(x) /4} + C.$

$\int _0^2 \left (x(x+1) + 1 \right ) \ dx = \answer {11/6}.$

$\int _0^\pi \sin (2x) \ dx = \answer {0}.$

Find the area under the graph of $y = \sqrt x$ from $x=0$ to $x=16$ .
The area is $\answer {128/3}$ .

Find the area under the graph of $y = 1/x$ from $x=1$ to $x=3$ .
The area is $\answer {\ln (3)}$ .

FTC, Part II

$\frac {d}{dx} \int _0^x \sin (2t^2) \ dt = \answer {\sin (2x^2)}.$

$\frac {d}{dx} \int _x^0 \sin (2t^2) \ dt = \answer {-\sin (2x^2)}.$

$\frac {d}{dx} \int _0^{3x} \sin (2t^2) \ dt = \answer {3\sin (18x^2)}.$

$\frac {d}{dx} \int _0^{x^3} \sin (2t^2) \ dt = \answer {3x^2\sin (2x^6)}.$

Implicit Differentiation

Find $\dfrac {dy}{dx}$ if $x^3 + y^3 = 6xy$ . $\frac {dy}{dx} = \answer {\frac {2y - x^2}{y^2 - 2x}}.$

Find $\dfrac {dy}{dx}$ if $y^4 = x^4 - xy$ . $\frac {dy}{dx} = \answer {\frac {4x^3 - y}{4y^3 +x}}.$

Related Rates

The radius of a circle is increasing at a rate of $2$ cm/min. How fast is the area increasing when the radius is $10$ cm? $\text {The area is increasing at a rate of} \ \answer {40\pi } \ \text {cm$^2$/min}.$

The radius of a sphere is increasing at a rate of $2$ cm/min. How fast is the volume increasing when the radius is $10$ cm? $(V= \frac 43 \pi r^3)$ $\text {The volume is increasing at a rate of} \ \answer {800\pi } \text {cm$^3$/min}.$

The short leg of a right triangle is increasing at a rate of $2$ cm/min and the long leg is increasing at a rate of $3$ cm/min. How fast is the hypotenuse increasing when the short leg is $12$ cm and the long leg is $16$ cm? $\text {The hypotenuse is increasing at a rate of} \ \answer {3.6} \text {cm/min}.$

Optimization

A farmer has 1200 feet of fence with which to fence in a rectangular pasture. Moreover, he would like to partition the pasture into two plots with fence parallel to the bottom side (horizontal). What dimensions will maximize the total area of his enclosed pastures?
The optimal length (horizontal) is $\answer {200}$ feet and
the optimal width (vertical) is $\answer {300}$ feet.

A fifth grader has $1200$ cm $^2$ of cardboard with which to construct a box with a square base and an open top. What is the maximum possible volume for his box?
The maximum possible volume is $\answer {4000} \ \text {cm}^3$ .

Max/Min, Inc/Dec and Concavity

Find the absolute maximum and the absolute minimum of the function $f(x) = x^3 - 3x+ 5$ on the interval $[0, 2]$ .
The absolute maximum is $\answer {7}$ occurring at $x = \answer {2}$ .
The absolute minimum is $\answer {3}$ occurring at $x = \answer {1}$ .

Find the absolute maximum and the absolute minimum of the function $f(x) = \sin (x) + \cos (x)$ on the interval $[0, \pi ]$ .
The absolute maximum is $\answer {\sqrt 2}$ occurring at $x = \answer {\pi /4}$ .
The absolute minimum is $\answer {-1}$ occurring at $x = \answer {\pi }$ .

The function $f(x) = x^3 - 27x$ is increasing on the interval(s):

$(-\infty , -3)$ and $(3, \infty )$ $(-\infty , -3)$ only $(3, \infty )$ only $(-3, 3)$

The function $f(x) = x^3 - 27x$ is concave down on the interval(s):

$(-\infty , 0)$ $(-\infty , 6)$ $(0, \infty )$ $(0, 6)$

Rectilinear Motion

A projectile is launched vertically. Its height (in feet) after $t$ seconds is given by $s(t) = 12 + 64t -16t^2$ .

The maximum height of the projectile is $\answer {76} \ \text {feet}$ .
The velocity of the object when it is $60$ feet up and falling is $\answer {-32} \ \text {ft/sec}$ .

A projectile is launched vertically. Its height (in feet) after $t$ seconds is given by $s(t) = 8 + 48t -16t^2$ .

The maximum height of the projectile is $\answer {44} \ \text {feet}$ .
The velocity of the object when it is $40$ feet up and rising is $\answer {16} \ \text {ft/sec}$ .

Mean Value Theorem

Find the value of ’ $c$ ’ from the Mean Value Theorem for the function $f(x) = 1/x^2$ on the interval $[1,2]$ $c=\answer {(8/3)^{1/3}}.$

Show that there is no value of ’ $c$ ’ from the Mean Value Theorem for the function $f(x) = 1/x^2$ on the interval $[-1, 1]$ . Why does this NOT contradict the statement of the theorem?

Area

Find the exact area under the graph of $y = e^{2x}$ from $x = 0$ to $x = 1$ .

Area under the curve $= \int _a^b f(x) \ dx$

$\text {area under the curve} = \answer {(e^2 -1)/2}.$

Find the exact area above the graph of $y = 1 - \sqrt x$ from $x = 1$ to $x = 4$ .

Area above the curve $= -\int _a^b f(x) \ dx$

$\int _1^4 1 - \sqrt x \ dx = \answer {-2/3},$ $\text {area above the curve} = \answer {2/3}.$