Emerging Scholars Sample Final Exam Question Solution Sketches

1. (a) f is negative when the graph is below the x-axis; this happens when -1.7 < x < 1.7 and when

5 < x < 6. (All values approximate!)

(b) f is increasing when it's going up; so 0 < x < 2.

(c) f is concave up when it's smiley! This happens when -3 < x < 2.

(d) This question is supposed to ask when the limit exists. The only place this

doesn't happen is when = 6 (since the right-hand limit is not equal to the left-hand limit there.)

(e) is continuous wherever you don't have to pick up your pencil, were you to draw it. This is

everywhere except x = 6 (again!)

(f) is differentiable wherever there are no sharp corners or cusps. This is everywhere,

except at =2, 4, and 6.

(g) The ' +3' moves the whole graph to the left 3 units, the '-' sign turns the whole graph upside

down, the '2' then makes all the heights twice as low (and the depths twice as deep on the

bottom side), and the '-4' shifts the whole graph down 4 units.

Here's a picture:

.

>

(h) To summarize: decreases until =0, so ' < 0 there.

increases until = 2, so ' > 0 there.

is then constant until = 4, so ' = 0 there.

is then linear with a negative slope of -2, so ' = -2 there.

is then constant until =8, so ' = 0 there until the end.

A graph:

(i) The graph of any antiderivative of (call it )

Summarizing: > 0 and decreasing until = -1.8, so is increasing and concave down there.

< 0 and decreasing until = 0, so is decreasing and concave down there.

< 0 and decreasing until = 1.8, so is decreasing and concave up there.

> 0 and increasing until = 2, so is increasing and concave up there.

> 0 and constant until = 4, so is increasing and linear there.

> 0 and decreasing until = 5, so is increasing and concave down there.

< 0 and decreasing until =6, so is decreasing and concave down there.

> 0 and constant until =8, so is increasing and linear there.

A graph:

2. To evaluate our limit graphically, graph the function and see what happens near =1. The graph:

As you can see the graph looks real nice near =1, in fact very close to the height of -0.5. So

graphically, my guess for the limit would be -0.5.

To do this limit numerically means to make a table of values by plugging in numbers closer and closer to = 1, and see if we can make sense of the answers. (This one I'll let you do!)

To do this limit algebraically, we need to do this with L'Hopital's rule, since just plugging in

the value =0 gives us something like - . So we get a common denominator, and find ourselves with , which is of the form , so we can L'Hopital our way through it. Taking the derivative of the top and of the bottom, we get which makes more sense if we simplify it to get , which still calls for L'Hopital's rule.

Taking the derivative of the top and of the bottom, we get , which we can evaluate and get . (All done!)

3. This is a Riemann sum problem! To find the lower bound, we look at each interval and choose the lowest rate we see, then multiply by the length of the time interval, as follows:

(10)(455) + 15(350) + 12(280) + 9(200) + 14(180) = 17,480.

To find the upper bound, we look at each interval and choose the highest rate we see, then multiply by the length of the time interval, as follows:

10(500) + 15(455) + 12(350) + 9(280) + 14(200) = 20,890.

So, no less than 17,480 gallons have flowed in, and no more than 20,890 gallons.

4. To approximate , we need to look at the graph and see if it's monotonic. The graph:

The graph is increasing from =0 to =1, then decreasing until =4. So, when we do our inequality, we have to do it twice: Once on the interval [0,1], and once on the interval [1,4], and for each interval we need to get a maximum error of 0.05 (the total error needs to be 0.1, so we are only allowed half of that on each interval)

So, on the interval [0,1], we have the inequality . Solving this for , we get , so that will work.

On the interval [1,4], we have the inequality . Solving this for , we get

, so that = 1557 is needed. (Ouch!) Please finish!

5. (a) Finding the derivative of by the definition means to write out

and simplify it. Getting a common denominator on top, we ge . Multiplying out the top gets us

. Cancelling out the 's and then plugging in =0 gives us our answer: .

(b) Finding the derivative of by the definition means to write out

and simplifying it. To do this we need to recall something from algebra: how to rationalize the numerator. In this case, we do that by multiplying the top and bottom by the conjugate of the top: . When we do this, we get the expression

. Multiplying out the top gives us

. We can then cancel the 's, and plug in to get:

.

6. Our function is, after simplifying to make the derivatives easier to take, .

(a) '( ) = . Factoring out , we get

'( ) = , which is 0 when = 5, and is undefined when = 0.

Plugging in test values, we find that ' is undefined when < 0,

' < 0 when 0 < < 5

' > 0 when > 5.

(b) ''(x) = + . Factoring out , we get

''( ) = , which is 0 when = 15, and is undefined when = 15.

Plugging in test values, we find that '' is undefined when ,

'' > 0 when 0 < < 15,

'' < 0 when > 15.

(c) We now need to find the limits: The furthest left the graph goes is to , since there's a square root involved. So we need to find . As approaches 0 from the right, the top gets very close to 5, the bottom gets very large. So the whole fraction gets larger without bound, and we say the limit = . The graph goes all the way to the right, so we also need to take the limit

. We could use L'Hopital's rule, but we could also say that the degree of the top is 1, while the degree of the bottom is 1/2 (as power functions). Since the higher the power, the faster the growth, the top grows way faster than the bottom, and the limit is again infinity.

(d) The graph:

7. The Gaspar the dog problem:

(a) The acceleration of Gaspar is the (instantaneous) slope on our graph. This is

greatest when = 0.

(b) As far as I can tell, Gaspar sees the dogcatcher at the 30 minute mark (he then reverses his direction!) The distance he ran until then is the area under his velocity graph until then, which is the quarter circle of radius 20 (so ) + area of the rectangle. (so 20(10) = 200). We get about 514.16, but the units are (m/hr)(minutes). To convert this to miles, we need to remember that 60 minutes = 1 hour, so we need to divide by 60, to get 8.57 miles

(c) To find Gaspar's average velocity we need to find , then divide by 60 minutes. Adding up the areas on top, and subtracting the areas on bottom, I get roughly 4.75 mi/hr.

(d) What is the difference between average speed and average velocity? Speed is never negative! In this case, just add up all the distances that Gaspar ran (remember, they're never negative), then divide by 60 minutes. I get roughly 13.75 mi/hr.

(e) Gaspar's average acceleration is defined to be , divided by 60 minutes. However, to evaluate this integral, we find an antiderivative of ( will do!), plug in 60, plug in 0, then subtract: In this case, according to our picture = = 0, so Gaspar's average acceleration is also 0.

(f) Gaspar only makes it back home if the distance he ran away from the home is equal to the distance he ran towards home. Poor Gaspar doesn't make it!

8. (a) Multiply out and use the power rule: .

(b) Substitute with , = , then use the power rule: .

(c) Substitute for each ot the last two, (or better, guess-and-check!):

.

(d) Guess-and-check all three: The first uses the power rule, the second the exponential rule for antiderivatives, the last you could substitute with , = . The final answer:

.

9. In order to find global minimums of a function, we usually start by taking derivatives. The Fundamental Theorem of Calculus tells us that = . Knowing when this is positive and when it is negative would be a real plus, so let's look at the graph of :

Remember, this is the derivative our function. So, our function is increasing, a lot, until is about 1.8, then decreases for a little while, then increases again. So, the lowest our function would ever be would be when = -1

10. We are given an expression for a rate (in gallons/hour), and are being asked for an accumulation (in gallons). The calculus way of doing this is to do an integral: . To evaluate this, we need to find an antiderivative of , plug in , plug in , then subtract. One such antiderivative is -10000 ; plugging in and , and subtracting gives us

(10,000 - 10,000 ), which is approximately 8646.647 gallons.

11. A good way of doing all of these acceleration-velocity-position problems is to try and find the equation of each, then rephrase the question in terms of the equations of each. Start with the acceleration: . Find the antiderivatives of both sides gives us , where is to be determined. We do know that the driver is doing 100 mph ( )* ( ) = ft/sec ) when We then get . Again taking the antiderivatives of both sides gives us

, where is the distance traveled after our driver has pressed the brake.

We know, however that when , the driver hasn't traveled any distance, so that = . We now want to know the answer to the question: What is when ? Well, when

(which is 70 mph in ft/sec). Solving this for gives us seconds. Then the distance traveled at that time is , approximately, feet. Sorry, it's ticket time!

The End!

Good Luck Everybody!