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1

Draw a graph of the following equation, R=4−12W. Plot W on the vertical axis and R on the horizontal axis.

First, we must recognize that to plot a function, the vertical axis variable (W) must be alone on one side of the equation. Here, we have the horizontal axis variable (R) alone. Thus, we must solve for W: R=4−12WOriginal functionR−4=−12WSubtract 4 from both sides8−2R=WMultiply both sides by âˆ’2

2

Draw a continuous function which begins at the origin, increases at a decreasing rate, reaches an inflection point, and then increases at an increasing rate. Show where each part of the function is concave and/or convex.

The function is concave from A to B, since a secant line between two points (red) lies under the function.

The function is convex from B to C, since a secant line between two points (blue) lies above the function.

The inflection point is at B, since the curvature of the function switches from concave to convex.

3

Solve the system of equations for x and y:

2x+y=204x−3y=10

We can use either the substitution or elimination methods. Here, I use substitution. Pick one equation and isolate one variable. Here, I pick the first equation and isolate x to one side: 2x+y=20Original equation2x=20−ySubtracting y from both sidesx=10−12yDividing both sides by 2

Now, substitute this in for x in the other equation, and solve for y:

4(10−12y)−3y=10Substituting 10−12y in for x40−2y−3y=10Distributing the 4−2y−3y=−30Subtracting 40 from both sides−5y=−30Simplifying y'sy=6Dividing both sides by âˆ’5

Now, substitute this into the first equation.

2x+(6)=20Substituting in 6 for y2x=14Subtracting 6 from both sidesx=7Dividingbothsidesby2

Now to double check, pick an equation and substitute both x and y to make sure the equation is true.

2(7)+(6)=20Substituting in 7 for xand 6 for y14+6=20Simplifying

We’ll verify again by checking the other equation is true.

4(7)−3(6)=10Substituting in 7 for xand 6 for y28−18=10Simplifying

4

Simplify the following equation:

Z=0.5X−0.5Y0.50.5X0.5Y−0.5

By the exponent rule for division (XaXb=Xa−b) for X and for Y:

Z=0.50.5X[−0.5−0.5]Y[0.5−(−0.5)]Z=(1)X−1Y1

By the rule for negative exponents (X−a=1Xa):

Z=Y1X1Z=YX

5

For the function f(x)=3x2+2x−7:

a.

Take the derivative of f(x), f′(x).

Use the power functions rule: Where f(x)=axn, f′(x)=(an)x(n−1). So for f(x)=3x2+2x−7, the derivative is (2∗3)x(2−1)+(1∗2)x(1−1) or 6x+2.

b.

In English, describe what the derivative of f(x) means

The function f′(x) is the derivative of function f(x), and it represents the rate of change (i.e. the slope) at every point defined on the function f(x).

c. 

Evaluate f′(4). In English, describe what this means.

Simply plug the value x=4 into the derivative function f′(4): 6(4)+2=26. The slope of the function f(x) at x=4 is 26.

6

Find the maximum value of:

f(x)=−2x2+16x

This is an unconstrained optimization problem for a single variable, so the maximum of the function occurs when its first derivative is equal to zero:

df(x)dx=0

So, by taking the first derivative, and setting it equal to zero, and solving for x:

df(x)dx=−4x+16=016=4x4=x

Going back to the original function, we find the maximum value of f(x) by plugging in x=4

f(4)=−2(4)2+16(4)f(4)=−2(16)+64f(4)=−32+64f(4)=32

7

a.

On a scale of 1 (worst) to 10 (best), rate your algebra skills (i.e. solving equations, graphing lines, working with fractions, etc.).

b.

Have you had any experience with calculus?

8

On a scale of 1 (least) to 10 (most), how anxious are you about this class? Feel free to elaborate any specific anxieties – it will make it more likely that I can speifically address them!