Practice for Exam Three

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Practice for Exam Three.

1 : Consider the function: $f(x) = \sage {p1fDisp}$ .

What is the sum of the $x$ values that are local maximums of $f(x)$ ? $\answer {\sage {p1ansMaxSum}}$
What is the sum of the $x$ values that are local minimums of $f(x)$ ? $\answer {\sage {p1ansMinSum}}$

Remember that local maximums and minimums occur where the derivative changes sign, not necessarily just where the derivative is zero.

2 :

Consider the function $\sage {p2fdisp}$ . Over what intervals is the function:

Concave Up: $\left (\answer {\sage {p2ansUpL1}},\answer {\sage {p2ansUpR1}}\right ) \cup \left (\answer {\sage {p2ansUpL2}},\answer {\sage {p2ansUpR2}}\right )$
Concave Down: $\left (\answer {\sage {p2ansDownL1}},\answer {\sage {p2ansDownR1}}\right ) \cup \left (\answer {\sage {p2ansDownL2}},\answer {\sage {p2ansDownR2}}\right )$

To determine concavity you need to use a sign chart on the second derivative. Where the second derivative is positive, the function is concave up. Where it is negative it is concave down.

3 : A particle’s distance traveled around a prototype accelerator at any time $t$ is given by $s(t)=\sage {p3f1}$ . Calculate the formula for the particles:

Velocity: $\answer {\sage {p3vel}}$
Acceleration: $\answer {\sage {p3acc}}$

Newtonian Mechanics were one of the first applications of calculus, and a primary motivation for their invention! Remember that the derivative of position is velocity, and the derivative of velocity is acceleration!

4 : Calculate what is needed to graph the function: $\sage {p4f1}$ .

Sum of the $x$ values that are domain restrictions: (If there are no restrictions, enter DNE) $\answer {DNE}$
Sum of the $x$ values that are local maximums (If there are none, enter DNE): $\answer {\sage {p4xmaxSum}}$
Sum of the $y$ values that are local maximums (If there are none, enter DNE): $\answer {\sage {p4ymaxSum}}$
Sum of the $x$ values that are local minimums (If there are none, enter DNE): $\answer {\sage {p4xminSum}}$
Sum of the $y$ values that are local minimums (If there are none, enter DNE): $\answer {\sage {p4yminSum}}$
Sum of the $x$ values that are points of inflection (If there are none, enter DNE): $\answer {\sage {p4xpoiSum}}$
Sum of the $y$ values that are points of inflection (If there are none, enter DNE): $\answer {\sage {p4ypoiSum}}$
Sum of the $x$ values that are zeros (If there are none, enter DNE): $\answer {\sage {p4xzeroSum}}$
The $y$ -intercept is: $(\answer {0},\answer {\sage {p4yInt}})$
The right horizontal asymptote is (if there isn’t one, enter DNE): $\answer {DNE}$
The left horizontal asymptote is (if there isn’t one, enter DNE): $\answer {DNE}$

Remember that, in order for a limit to exist, both the left and right limits must exist and be equal.

5 : Use the function $f(x) = \sage {p5f1}$ at $x=\sage {p5anchor}$ to approximate the value of $\sqrt [\sage {p5pwr1}]{\sage {p5target}}$ . $\answer {\sage {p5ans}}$ .

Remember, linear approximation is nothing more than building a tangent line at the given (nice) $x$ -value, and then evaluating that tangent line formula at the (not-nice) $x$ -value you want to approximate!

6 :

A $\sage {p6ladder}$ foot ladder is resting against the wall. The bottom is initially $\sage {p6base}$ feet away from the wall and is being pushed towards the wall at a rate of $\sage {p6fall}$ ft/sec. How fast is the top of the ladder moving up the wall $\sage {p6time}$ seconds after we start pushing? $\answer {\sage {p6ans}}$

It’s always good to draw a picture; the ladder in this case forms a nice right triangle which probably triggers a Pavlovian like reflex to use Pythagorean’s theorem. Lean into this reflex, it will help you!

7 : We want to construct a box whose base length is $\sage {p7boxmulti}$ times the base width. The material used to build the top and bottom cost $ $\sage {p7costtop}$ per square feet and the material used to build the sides cost $ $\sage {p7costedge}$ per square feet. If the box must have a volume of $\sage {p7vol}$ cubic feet determine the dimensions that will minimize the cost to build the box.

Width: $\answer {\sage {p7Width}}$ feet
Length: $\answer {\sage {p7Length}}$ feet
Height: $\answer {\sage {p7Height}}$ feet

When in doubt, draw a picture! Note that the Length should be the dimension that is $\sage {p7boxmulti}$ times larger than the width dimension. Remember it is typically the case that the constraint equation (the set volume here) is used to remove a variable in the extrema equation (the cost equation in this case).