Math 006 - Quizzes

The mid-term exam will be closely based on the weekly quizzes.

I am going to put up a list of weekly sample quizzes, and worked solutions on this page. You can also look here before the day of the quiz, to check you've got the right way to do the question.

• Quiz 3
• Quiz 4
• Quiz 5
• Quiz 6
• Quiz 7
• Quiz 8
• Quiz 9
• Quiz 10

First Quiz

Algebra Review

The first quiz was basic algebra, as in the first few pages of the text book.

Second Quiz

Calculation of derivative as a limit

For a brief example, see page for week 3.

Finding equation for tangent at a point

Question:

Find the equation for the tangent to y=x³-x at x=2

Answer:

Sketch of what the question is about:

In the digaram, the axes are in black, the curve is in red, and the tangent is in blue.

The tangent is a straight line, so it's equation is something like:

The derivative of the function at x=2 is the slope (this is the definition of derivative).

To find the slope, we could use the method of finding derivative by tables, or we can use formulars. Since the formulars are quicker, we'll do it like that.

We use "derivative of difference is difference of derivatives"

This means d/dx(x³-x) = d/dx(x³)-d/dx(x)

Next use "power rule", ie d/dx(xⁿ=nx^n-1, which means that d/dx(x³) = 3x².

Also use the fact that the derivative of x is 1, (since it's a line, with slope 1),

putting these together, we get

So when x=2, the slope is 3*2² -1=3*4 -1 = 11

So the equation of the tangent is

To find b, we substitue the values (x,y)=(2,2³-2) = (2,6)

So

so b = 16. So the equation for the tangent to the graph of y=x³-x at x=2 is

Fourth Quiz

Find a value of x where the tangent to y=x²+3x+1 has slope 5

Answer

The derivative of y=x²+3x+1 is 2x + 3.

Where is the slope of the tangent 5? This will happen when the derivative is 5, ie, when 2x + 3 = 5. The solution is x=1.

Note, for this question, the above is the final answer. But sometimes in a quiz or on an exam, eg, the midterm, or on assignments, you are asked to sketch a graph to go with the answer, eg, sketch a graph of the tangent in this case. If you are asked to sketch a graph, please include the following:

• Lable the x and y axes
• Put some indication of scale on the axes
• Lable what you are drawing, eg, write the equation of the curve and the equation for the tangent or secants drawn.

Also, it it generally best to use a sharp pencil, (or if you use a pen, make sure you are careful not to make mistakes, and use a fine enough point that the diagram is clear) and a ruler for straight lines. Using something like quad ruled paper (graph paper) is also useful for getting a more acurate graph.

Also it is a good idea to include a table of some values of (x,f(x)) you plot, to show how you plotted it, or some other justification for what you are drawing.

Example:

Question

(i)Find the critical values of the function 6x⁵ -10x³
(ii)What are the absolute maximum and minimum values of this function on the interval [0,2]?

Answer

(i) The critical values are places where the derivative of the function is either 0 or undefined. This function's derivative is always defined, it is given by

Where is this equal to zero? We need to solve

Which we do by factorixing:

So the solutions are x = 0 or x = -1 or x = 1. This means the critical values of the function are -1,0,1.

(ii) To find the absolute maximum and minimum values, we need to find the values of the functions at the critical points and at the end points of the interval, and see where is it greatest and least, ie, complete the following table:

So the absolute maximum value of this function on the interval [0,2] is 112, which is attained at x=2, and the absolute minimum value is -4, which is attained at x=1.

Sixth Quiz

(a) What is the horizontal asymptote of the function (2x+1)/(x+1) ?

(b) What is the vertical asymptote of the function (2x+1)/(x+1) ?

(c) State whether this function is concave up or down on the interval [2,infinity].

Answer

(a): The horizontal asymptote tells us what happens as x gets infinite (not all functions have a horizontal asymptote).

Well, what happens for (2x + 1)/(x + 1) as x gets very big?

Try some examples:

Well, you can see these numbers are getting closer to 2, so:

We have to check about what happens when x gets large and negative too, but the same thing happens, eg:

Note, in limit notation, the above is written as

(b): The vertical asymptote tells us where the function might equal infinity. The only way (2x + 1)/(x + 1) can be infinite is if (x + 1) is zero. This means x = -1; so

(Aside: strictly speaking, this means the function's not really defined if (x + 1)=0 ; that's why we go into all that stuff about limits, to describe what we mean by infinity; that is, infinity means "very very big, and the closer (x + 1) is to zero, the bigger ("just about anything" divided by (x + 1) is.)

Numerically, we have for example:

Note, in limit notation, the this is written as

This tells us that as x gets close to -1, but a bit bigger than 1, f(x) gets close to , and as x gets close to -1, but a bit less than 1,

Now, the answers to the question are given in the boxes above, but suppose you wanted to put this information on a graph - how would it look?

The lines just look like this:

You have to fill in the function in one of the following ways, roughly. Which might be the right way?

Note that as x gets big, the values of f(x) are always a bit less than 2, and for x big and negative, the values are bigger than 2. Also we saw what the limits were as x tends to -1 from above and below -1; so that means we can say that the shape of the function is something like this:

(c). We want to know about where the graph is concave up or down. (click here to see the definition of concave up and down if you forgot). Well, if our diagram is correct, then it shows that the function is concave down.

But we've not really got any justification that our function does not look like this:

The most sure way to check whether the function is concave up or down is to find the second derivative, as follows:

If x > -1, then (x+1) is positive, so so is (x+1)³, so so is 1/(x+1)³; but the second derivative has a -2; so this means that for x > -1, f''(x) is negative, so this means the function is concave down on the interval (-1, infinity), and so also on the interval [2,infinity).

Seventh Quiz

Question: Find the intercepts, asymptotes, local max and min, of the function (x² - 5)/(x -3), and use this to sketch the graph on the interval [-6,6].

This question asks you to calculate three things to help sketch the function, and then you use this information to sketch it.

Note, we have covered more aspects in curve sketching than just these three, but in an exam, or on a test, in this course, you will only be asked to calculate at most three things in drawing the graph. Actually, these three things in this case are enough to get a pretty good idea of what the graph looks like.

There are several things to calculate for this question. The things to make sure you state clearly are the things in the boxes below. There are several parts of this question:

• (A). Intercepts
• (B). Asymptotes:
• (C). Max and Min:
• (D). Graph:

(A). Intercepts:

y intercept: This is when x = 0:

So

x intercept: This is when y = 0: So, when does

To get this, you need

Here's a picture on a graph to show where the intercepts are:

(B). Asymptotes:

Horizontal Asymptote: This is a horzontal line that the funtion should get close to for big values of x:

So to find where this line is, we have to see what f(x) looks like for large values of x.

This is because in comparison to 5 and 3, x is very big, so 5 and 3 make hardly any difference to the size of f(x) for large x, so the function is close to x. So this means there is no horizonal asymptote; as x gets bigger and bigger, the function gets bigger and bigger - or bigger and bigger but negative for negative x.

What we get is that the funtion gets closer to this red line for very large x. (This is covered in the section on slant asymptotes in the text book, which we will not cover in the course, so you don't have to find the equation of this line):

Vertical asymptotes: This means a vertical line that the function may get close to somewhere:

To find this, we have to know if the function is infinite anywhere, so can

This can only happen if the bottom of the fraction is zero, ie, if x = 3. So this is the vertical asymptote

(C). Max and Min:

To find the maximum and minimum on the interval [-6,6], we need to find where f '(x) = 0. The derivative is found using the quotient rule:

for this to be zero, we need the top of the fraction to be zero, ie

If we factor this, we get

So the critical values are at x = 5 and x = 1.

We can calculate a table for the values of f(x) at the critical values, and at the end points of the interval [-6,6]:

Which of the critical values is maximum, and which is minimum? You can make certain these are local max and min by finding the second derivative, or by finding f(x) for x close to 5 and 1; so from the data we already have, we can be pretty sure that f(5) is a local minimum, and f(1) is a local maximum. To be sure, find f''(x) (actually, there are three ways to check whether it's a local max or min - as done in class - examine f(x) near x=1 and 5, examine f ' (x) near 1 and 5, or find f ''(1) and f ''(5)):

We already have

So

We have

and

So, since f ''(1) is negative, it must be a local maximum, and since f ''(5) is positive, it must be a local minimum.

(Note, that here I've given the Local max and min, not absolute max and min, which was what was done in quiz 5 above)

We can add all of this information to the graph:

Sketch of the graph

Finally, to complete the question, all we need to do is draw a line which goes through the points we've found above, it must get close to the vertical asymptote at x = 3, and it has a local maximum at x = 1, and a local minimum at x = 5.

The line we draw should look like this:

Eighth Quiz

Question: Find the derivative of the function

Use this to find the equation for the tangent to f(x) at x = 0

Answer: Remember, there are two steps for finding the equation of the tangent of f(x) at a point with x = A

Finding m

This is exactly the same proceedure we've seen before.

So, find the derivative of f(x) = sin(x² + x + 1)

We're going to use the chain rule, since this is a function of a function; compare it with (x² + x + 1)³; there is an outside part, the sin(?) or the ?³, and the inside part, y = (x² + x + 1).

The derivative of of sin(y) is cos(y)

The derivative of (x² + x + 1) is (2x + 1)

Put these together, to get the derivative of the whole thing:

So at x=0, the slope of the tangent is

Note, to find cos(1), make sure you set your calculator to radians mode. Always assume work with angles is in radians mode when you're doing calculus with trionometry

Finding b

When x = 0, f(0) = sin(1) = 0.84 So substituting in y = mx + b, we have

So b = 0.84, and

Quiz 9

Question Find the critical values of the function

Answer

The derivative of this function is:

When is this zero? Either x =0, or

This can be solved using logrithms:

So

So the critical values are x=0 and x=1.5448 and x=-1.5448.

Quiz 10

Question: The population of a certain city grows at 10% per year. The population in 1990 was 200 000. What will the popultion be in the year 2000? (Assume the rate of growth remains constant.)

Answer: For this question, we assume we have exponential growth, that is, we assume that the function giving the population of the city is of the form

We have two unknowns, C and k, and two pieces of information, which will allow us to find C and k:

This means:

We substitute both these peices of information into the equation for f(t):

We know that e⁰ = 1 (for any number a, a⁰=1). So

The way to solve an equation of the above form is to take logs of both sides - use the natural log, ln = log_e, which is the inverse to the exponential function e^x; ie, ln(e^y) = y, and e^ln(y) = y.

So we have:

So we have

Then you can use your calculator to find k:

So the equation for the population is

The question asks "what is the population at time t=10 ?" (10 years after the start time of measuring). We have:

Note, we've finished the question now, but we could do some more algebra - and the following is worth while knowing:

We had

There is a rule for logarithms which says that ln(a) - ln(b) = ln(a/b), so we have:

Now we have

Remember that ln and e are inverse, so e^ln(1.1) = 1.1, so we have

And we can use this to find f(10), and it will give the same answer as before. You might ask "where does 1.1 come from?" It is "1 + 10%". If the population had been growing at 17% per year, we would have "1 + 17%" appearing. 17% = 17/100, so 1+17% = 1.17, and the equation would be

where P is the starting population, eg, if we started at 500 people, growing at 17% per year, we get

You should notice that this is the same kind of formula as that for compound interest given in class.

And if you remember the formular you can skip the algebra - but if you don't like memorizing formulae, if you understand the algebra you can reconstruct the formular.