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 8 (Log and Exponential Derivatives) and Lesson 9: Curve-Sketching from my Intro Calculus book to prepare for this assignment.

Study Lesson 8, #1 to prepare for this question.  Their part (d) is similar to my #1(p).  Especially make sure you have learned the var-to-var type of problem as illustrated in my #1(n), (o) and (p). 

Do note that you can OMIT Lesson 8, #5 and #6.  That inverse derivative stuff is not something they cover anymore.

Similar to my Lesson 8, #2.

The rest of this assignment should be attempted after you have studied Lesson 9 in my book.

This question is asking for the critical numbers.  That means they want the critical points and singular points.  All the points where the derivative is either zero or undefined.  That is the top and bottom zeros of the first derivative are the critical numbers (if the bottom zero is a vertical asymptote, it is not a critical number). 

Hint for part (b): Rewrite |7x-4| as sqrt[ (7x-4)^2 ] in order to compute f'(x).

Similar to my Lesson 9, #5.  Make sure you include the sentences I box in in your answer as that is necessary to justify your conclusions.  But, this is not a polynomial!  However, a fifth root function is always defined since you can fifth root both positive and negative numbers.  Therefore, this function is continuous since it is a fifth root function times a polynomial.

Personally, I would avoid product rule by rewriting the fifth root of x as x^(1/5), then multiplying it through the brackets and adding exponents to simplify.  You will get a negative exponent when you do the derivative.  Pull that down to the denominator and then get a common denominator to properly identify the top and bottom zeros.  You may find it easier to just guess at what the top and bottom zeros are.  Be organized.  Sub in x=0, 1, -1, 2, -2, ...  There is no shame in just using trial and error to find zeros when things are difficult to factor.  Facts are facts.  If you find an x-value that causes either a top or bottom zero, then there is no doubt. 

Hint: There are two critical numbers in [-1, 32], one critical point and one singular point.

This is a Mean Value Theorem question.  Click the link below for the procedure to follow to "verify" the Mean Value Theorem:

Be sure to point out that the given function is a rational function that is only undefined at x= -4, and so it is certainly continuous on the closed interval [-3,0], as required by Mean Value Theorem.  After you compute the derivative, f'(x), it should be obvious that the derivative exists everywhere but x=-4, too (that will cause zero in the denominator, so derivative does not exist at -4), and so f(x) is differentiable on the open interval (-3,0), therefore the Mean Value Theorem applies.

Hint: You should get two answers for c, but only one of them will be within [-3, 0].  Show both answers, and explain why one is discarded.  The final answer should be, something like, "Since c= (blank), we have verified the Mean Value Theorem for this function on [-3,0]."

A classic curve sketch problem. Make sure you have done my Lesson 9, #3 and #4, and Practise Problems 1-16 in order to be thoroughly prepared for this problem.  This one is quite similar to my Lecture Problem #3(b).

Since you are asked to show all your work, I suggest, when finding the vertical and horizontal asymptotes, that you actually compute the k/0 limits and solve the infinity limits formally, as I teach in Lesson 2, to ensure you are not penalized. 

You should perhaps also prove whether the function is even, odd or neither (although you will already know the truth by looking at your graph), by computing and simplifying f(-x) and comparing the answer to the original f(x). 

See my tips for question 6 above.  This one is quite similar to my Lecture Problem #4.

Hint: Don't let that x^2-6 distract you in f"(x).  Certainly, that is a difference of squares, but, it is not a factor of the top.  It is a term.  You are given (x^2-6)^2 + 60.  In other words, you are given (something)-squared + 60.  That is the pattern u^2 + 60.  But, as I say right at the start of Lesson 9, x^2 PLUS a number, does not factor, has no zeros.  Who cares if x^2-6 is ever zero? That is not going to make (x^2-6)^2+60 become zero.  It would just be 0+60 which is 60.  And, if x^2-6 is anything other than zero, it will be squared, and be a positive number, adding to 60, giving you a value even bigger than 60, and so certainly not 0.

Never forget! (something)-squared PLUS a number can never be 0. It has NO ZEROS, it is always positive.

That's why the prof wrote it that way.  You could waste your time squaring the x^2-6 then adding 60 to make a polynomial, but that polynomial would be hard to analyze (it would be an x^4 polynomial), and after any work and analysis you did, you would simply discover it has no zeros.