Final Exam Prep - 1st Semester

General final exam studying help: 

Our final will feel very much like a regular chapter test, except it will be on material from the entire semester. This could be good (easy stuff like limits from Chapter 1) or tough (difficult stuff like mean value theorem from Chapter 5). One good thing to keep in mind is that you’ve been consistently working on material from September and October within the context of the problems we’re learning right now. So, hopefully the older material won’t feel like you’ve totally forgotten it.

While you’re doing the review packets, keep this in mind…

You get the problem right on the first try, without looking at notes or anything.

• This means you understand how to do this one! Move on to the next one.

You get the problem wrong on the first try, but figure out how to do it after looking at the answer key.

• This means you need to keep working on this type of problem. Find another similar one to practice it, or have me or someone else make a new one up for you.

You get the problem wrong, and can’t figure out how to do it, even after looking at the answer key and your notes.

• Ask someone for help ASAP! That means calling or texting a friend, or emailing me this evening.

I hope this helps! Let me know what questions you have!

Final Exam Review Packet help:

11) You could use any definition of a derivative formula. Usually, people find it easiest to use the forward difference quotient:

lim    f(x + h) - f(x)
h->0        h

So, we need to find f(x) and f(x + h):

f(x) = x^3

f(x + h) = (x + h)^3

We can plug this in to the derivative formula, then begin simplifying:

lim    (x + h)^3 - x^3
h->0        h

Write out (x + h)^3 three times, then distribute. I would distribute two of the factors together, then distribute the third. Here are the steps:

(x + h)(x + h)(x + h)
(x + h)(x^2 + 2xh + h^2)
x^3 + 2x^2h + xh^2 + hx^2 + 2xh^2 + h^3
Then combine like terms: x^3 + 3x^2h + 3xh^2 + h^3

Here’s what your derivative formula looks like now:

lim    x^3 + 3x^2h + 3xh^2 + h^3 - x^3
h->0                h

The x^3 and -x^3 cancel out:

lim    3x^2h + 3xh^2 + h^3
h->0            h

Then, we can cancel an h from every term, because it’s common to all the terms on top and bottom:

lim    3x^2 + 3xh + h^2
h->0  

Now, we can finally find the limit as h approaches 0. Plug in 0 for each h:

3x^2 + 3x(0) + (0)^2

= 3x^2

Which is what we would get from doing the plain old power rule! :)

12) The limit of a sum property says that you can do this:

lim    x^2 + 6x + 10
x—>1

and get an answer of 17, rather than having to split it up like this:

lim    x^2    +    lim    6x    +    lim    10
x—>1               x—>1             x—>1

and get an answer of 1, an answer of 6, and an answer of 10, then add them together to get 17.

This is not a property that I’ll test you on specifically! The question is just meant to make you think about limits and how to find them. :)

17) This wants you to use the quotient rule to find the derivative of tan x. We can do this, because we know:

tan x = sin x / cos x

Here’s the set-up for the quotient rule:

u = sin x
u’ = cos x

v = cos x
v’ = -sin x

Quotient rule:

cos x*cos x - sin x*-sin x
        (cos x)^2

But, cos x*cos x is the same as saying (cos x)^2. The two negatives in the second term combine, so we can rewrite as:

(cos x)^2 + (sin x)^2
    (cos x)^2

Then, we need to recognize the numerator as one of our fundamental trig identities: (sin x)^2 + (cos x)^2 = 1. Now our equation simplifies to this:

      1        
(cos x)^2

Then the reciprocal of cos x is sec x, which gives us the answer of:

(sec x)^2

29) This one is tough to explain in writing, but here’s my best shot:

dx means the same thing as “change in x”, or “delta x”. Draw a Riemann Sum-like rectangle on the graph (however wide and wherever you want), and label the width of it dx. That’s part of what a Riemann Sum rectangle shows: its width shows a change in x from one x-value to the next x-value.

When you label a point (x, F) in that rectangle, that is supposed to help show you that the height of your rectangle is F.

Write an equation for the area of that rectangle:

height is F * width is dx

= F * dx

So, your equation is:

dW= F dx

The dW is the derivative of W. Think about what you would have to do to each side of your equation to make it an equation with regular W: you’d have to integrate each side (because an integral is the opposite of a derivative):

dW = F dx

(integral sign) dW = (integral sign) F dx

W = (integral sign) F dx

Look, it’s like a regular integral set up! You have an equation F with a dx next to it inside of the integral sign. This is because every single integral we work with is just the area of a rectangle, where the equation is the height of the rectangle and the width is dx.

31) It's not a proof that you need to know for the final exam tomorrow. Notice that it only wants you to “sketch a graph” to show that you understand the conclusion of the mean value theorem. So, your graph should be continuous and differentiable, with an interval from x = a to x = b with a secant line connecting a and b. Then, find an x = c (and label it) where there’s a parallel tangent line (and draw the tangent line). This would be a complete answer.

Bicycle #2) Remember, the sign of velocity indicates direction. So, the cyclist is changing directions whenever velocity changes from negative to positive or vice versa. This happens on our v(t) graph at t = 3 only.

Bicycle #3) In the intro to the problem, it says that “east is considered the positive direction”. So, the cyclist is traveling east whenever v(t) has positive values. This happens on our v(t) graph in the interval [0, 3). t = 0 is included because it says that the cyclist is starting at a velocity of approximately 42mph. t = 3 is NOT included, because the cyclist is traveling 0 mph at that moment (which means stopped, and not traveling anywhere).

Bicycle #5) Because the given graph is v(t), the derivative of it will be acceleration a(t).  This problem wants you to find the "average acceleration", which is the same as the average derivative, which is the same as the average slope.  In other words, you need to find the slope from x = 0 to x = 9.  Estimate the y-values for both points, then find the slope!

Bicycle #6) Because the given graph is v(t), the derivative (the slope) of it will be acceleration a(t).  So, this question is asking you to estimate the slope of the tangent line at x = 5.

Bicycle #8) This will also be an estimate.  This question is the same as asking: "Where is the instantaneous slope the same as the average slope?"  You found the average slope in Problem #5, so now you need to estimate where the instantaneous slope would be the same.  To do this, estimate the two locations where the tangent line would be parallel to the secant line from x = 0 to x = 9.

Here’s more of an explanation:

One of the things to keep in mind when you’re looking at a velocity graph or equation is this:

Derivative (slope) of displacement = velocity
Derivative (slope) of velocity = acceleration

So in this problem, when they’re asking about “instantaneous acceleration” and “average acceleration”, they mean “instantaneous slope” and “average slope” for this v(t) graph. In other words, they’re asking you to apply Mean Value Theorem! Average slope is finding the slope of the entire interval (the secant line slope) and instantaneous slope will be finding the derivative at a certain point.

To actually do this problem, where you’re not given an equation, you will need to draw lines and estimate. Start by drawing the secant line from x = 0 to x = 9. Then, estimate the other two points on your graph where the tangent line would be parallel to this secant line.

To draw the graph of a derivative) On your given graph, any max or min (i.e. a point with a slope of 0) will be a zero on the derivative graph.  Remember, the slopes on your given graph will be values on the derivative graph.  So, if you're given a segment with positive slopes, then your derivative graph will have positive y-values at those x-values.

Video: Final review #18 (Semester 1)