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Don't have my book or audio? You can download a free sample of my book and audio lectures containing Lessons 1 and 2: Did you read my tips on how to study and learn this course? If not, here is a link to those important suggestions: Here is a link to the actual assignment, in case you don't have it: Study Lesson 1 (Skills Review), Lesson 2 (Limits), Lesson 3 (Continuity), and Lesson 4 (Definition of Derivative) from my Intro Calculus book to prepare
for this assignment.
Of course, always seek out assistance from my book, your course notes, etc. if you ever hit a question you don't understand, but try not to be learning things as you do an assignment. Learn first, then put your learning to the test. This question uses the skills that I teach in Lesson 4. It is actually the definition of derivative but without doing the limit as h goes to 0. I suggest you study that lesson to learn the necessary skills (such as simplifying Triple Deckers). Do not
use limits, of course, but make sure you have simplified to the point of factoring h out of the top and canceling with the h below. Also note that the a in their questions is just like the x in my examples. You are subbing a+h in place of x in the function, and a in place of x.
NOTE: They are NOT saying there will be no h in your final answer. They are just warning you that you will be able
to get rid of some of the h's. As I teach in Lesson 4, you will always be able to factor h out of the top and cancel with the h in the bottom if you have done your algebra properly.
In other words, do exactly what I do in my Lesson 4, #1(b) and #2(a) and (b), but never even mention the limit as h approaches 0, and, therefore, don't sub h=0 in at the end. And use a instead of x. You should be able to simplify
the numerator in each problem and factor h out to cancel with the h in the denominator. Once you cancel the h factors in the numerator and denominator you are done (since you are not doing the limit as h approaches 0).
Hint for part (c):
2 to the power of "a+h" can also be expressed as 2 to the power of "a" times 2 to the power of "h" by exponent laws (when you are multiplying the same base, you add the
exponents). That will enable some factoring at least, but you won't be able to cancel an h in this one (note that the question ignored this part when talking about h "not appearing."
Be sure to study my Lesson 1, pages 5-20 to prepare for this question. You may find my audio on making sign diagrams helpful here to properly understand what I am teaching you about how to make a sign diagram. I have given you the entirety of my audio for
Lessons 1 and 2 in the free samples above. Here is my audio introducing sign diagrams:
Part (a) You can google complete the square if you want to be reminded of how to convert this given quadratic equation into the parabola form they request, but I don't think that you will ever have to do something like this again. The
h and k they are referring to are the x and y coordinates, respectively, of the vertex of the parabola. In general, for any quadratic equation (ax^2 + bx + c), the x coordinate of the vertex will be -b divided by (2a). Which is to say, h = - b / (2a). Once you know h, you can sub it into the given quadratic equation to compute k, the y-coordinate of the
vertex.
Once you know h and k, sub them into the given format and you have answered part (a).
Part (b)
Just make a quick table of values. Plot the vertex you just found, and choose one or two other points on each side of the vertex.
Part (c)
The vertical line test checks if a graph is a function or
not. If a vertical line can never pass through more than one point on the graph, the graph represents a function
The horizontal line test checks if a graph's inverse would be a function. If a horizontal line can never pass through more than one point on the graph, then its inverse would be a function.
A graph is a one-to-one function if and only if it passes both the vertical and
horizontal line tests. That proves that for any one value of x there is only one value of y on the graph. The function is one-to-one.
Sketch the graph of the parabola, and it is obvious that it fails the horizontal line test. So you must restrict the domain to make it one-to-one. The most logical thing to do is to restrict the domain to all the x values starting from the vertex and to the
right.
Part (d)
Be sure to use the (h, k) form they had you generate in part (a). Much easier to do the algebra with that one.
I show you how to find the inverse of a function in Lesson 8 of my book (just after #4 in the lecture, page 230). I like to immediately have x and y change places, then proceed to isolate
y.
Once you have isolated y, make sure you replace y with f^-1 (x), the f-inverse symbol.
Since an inverse changes the x and y around, it is sort of rotating a graph sideways (and flipping it). What was vertical becomes horizontal, and vice-versa. Thus, a graph that passes the horizontal line test, will have an inverse that passes the vertical line test, proving that the inverse will be a function. Only
one-to-one functions will have inverses that are also function.
Be warned! If you are doing your math right, you should get two answers for the inverse (which is impossible) because you will have to take a square root. WHENEVER YOU HAVE TO USE A SQUARE ROOT TO HELP SOLVE A PROBLEM, YOU MUST REALIZE THERE ARE TWO POSSIBLE ANSWERS: THE POSITIVE SQUARE ROOT OR THE NEGATIVE SQUARE ROOT. However, only one of those answers fits your
restricted domain from part (c). Show both answers, then explain why you are discarding one of them.
For example, if you were solving the equation x^2 = 9 (nothing like this particular problem), you would square root both sides and get x = + square root of 9 or x = - square root of 9. So x= 3 or x= -3. Make sure you have read the Logs and Exponentials section of Lesson 1 in my book (starts on page 23).
You may also find my audio working through my Lesson 1, #4 helpful (included in my free samples
above): Part (a) Use a log law to combine the left side into one logarithm, then convert the log to an exponential. Be careful! Be sure to check your answer to see if it works in the original equation because you can only do the logarithm of positive
numbers. Note that it is not important whether a solution is positive or negative. Rather, does subbing the solution into the original equation cause the log of a negative or log of 0? Make sure you discuss all this or you will be penalized. You can never assume the solution to a log equation is correct. The algebra can be fooled. Solutions must be checked in the original equation to ensure that you are always getting
log(positive).
Part (b) Think carefully. If you square both sides, what does that do to the exponential on the left? Remember if a power is raised to a power, you multiply the powers. For example,
(x to the power of 2) all raised to the power of 3 is x to the power of 6.
Then, you have to use logs here (use ln, the natural log). Challenging
algebra. Once you have used logs and simplified by using log laws (watch your brackets), collect all the terms with x to one side, and factor x out in order to isolate it.
Part (c) This is actually a quadratic equation. For example, if e^x is t, then e^(2x) is t^2 (t-squared). Solve for t first. Once you know what t is (there are two answers), solve for x since t is e^x. But one of those answers
will have to be discarded!
Part (a) To sketch each piece of this function, merely plot 2 or 3 points for the domain of each piece and connect the dots. For example, graph y= 3x for the region of (0, 1) by plotting 2 points in that
region. For sure, those points should be the endpoints at 0 and at 1. However, since that region is up to but not including the endpoints, plot a "hole" at that location rather than a dot. Which is to say, there will be a hole at (0,0) and another hole at (1,3) and those two holes are connected by a line representing y=3x in that region.
Be sure that you always plot the endpoint(s) for each piece. If the domain of a piece includes the
endpoint, plot a dot. If the domain is up to but not including the endpoint, plot a hole at the y value that it almost reaches.
Part (b) Once you have graphed all five pieces of this function, it is a simple matter to read off the solutions for all the limits, similar to what I do with my opening example in Lesson 2 of my book. You might find my audio discussing the start of Lesson 2 helpful here.
That is included in the free sample at the top of this message.
You could also solve the limits algebraically using the procedure I show in Lesson 3, but the fact that they have asked you to draw the graph suggests they intend for
you to merely read the limits off of the graph, no work. You can use the algebraic method to check your answers though.
Remember: It is irrelevant what is AT x=a when doing the limit as x approaches a. - For x approaching a-, find the piece of graph that is immediately to the left of a, and find what y-value that piece connects to (the endpoint of that
piece, be it a hole or a dot, is the limit).
- For x approaching a+, find the piece of graph that is immediately to the right of a, and find what y-value that piece connects to (the endpoint of that piece, be it a hole or a dot, is the limit).
- If and only if the limits for a- and a+ agree, the limit approaching a exists and has that
value.
Part (c) You can answer this question by merely observing the graph you have drawn and saying whether or not the graph is broken or has a hole in it at the x-values in question. If you want to be thorough, you can mention the definition of continuity to justify your answer. This is a good run-through of limits. Study Lesson 2 thoroughly to prepare for this question. This is the most important question on this assignment, in my opinion. Many of these limits could appear on your final
exam.
Part (a) As always when solving a limit, sub in the given x and see what happens first. Only once you know which of the three possibilities you are dealing with, do you know what to do next. Hint: You won't have to do much next.
Part (b) A K/0 limit problem like my #6 and #7.
Part
(c)
Classic 0/0 limit problem like my #1. However, the x-cubed polynomial on top makes it difficult to factor this expression. You should have realized that 2 is a zero on top and bottom, so (x-2) is a factor on top and bottom as my factoring tip in Lesson 2 discusses. This will require polynomial long division. You may want to Google that term to get an idea how to do this. Here is a video I found on YouTube
that explains things well:
In your case, you need to divide x-2 into x^3 + 2x^2 - 5x - 6 to see what (x-2) multiplies with to factor the top properly. Check your answer. If you have factored the top properly, you should be able to multiply it back together to get x^3 + 2x^2 - 5x - 6.
Part (d)
Standard
conjugate problem like my #2 and #3. Watch your bracketing!
Part (e)
Standard absolute value limit like my #4 and #5.
Part (f) Be careful! This is a 0/0 limit but it then turns into a K/0 limit, so you have to then deal with that.
Part (g)
This one is k/0 again.
Can you come up with a good argument for the sign of the bottom?
This was obviously supposed to be a continuity question like my Lesson 3, #1 to 3. But, he forgot to give you the function! Either contact the prof to get this typo fixed, or stay quiet and the question will have to be
cancelled.
This uses the Intermediate Value Theorem like my Lesson 3, #4 and 5. Prove g(x) has at least one zero on (4,5). Remember, that if the sign changes, you know there must be at least one zero. You can never know from this theorem alone
exactly how many zeros there are. Make sure you state that g(x) is continuous (why is it continuous?). Be sure that you say "by Intermediate Value Theorem" as your justification.
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