Math 1500: Tips for Assignment 3 (includes a copy of the assignment)

Published: Sun, 11/13/16

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Did you miss my Tips for Assignment 1? Click here.
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The department posted solutions for Assignment 1 (these are not my solutions). Click here.
Tips for Assignment 3
Here is a link to the actual assignment, in case you don't have it handy:
Study Lesson 8 (Log and Exponential Derivatives) and Lesson 9: Curve-Sketching from my Intro Calculus book to prepare for this assignment.
Question 1
Similar to my Lesson 8, question 1.  Make sure you have studied my Lesson 8, questions 1(n) to (p) before attempting part (d).
Question 2
Similar to my Lesson 8, question 2.
Question 3
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)

Make sure you give both the x and y coordinates of your critical numbers, even though it would be fine to just give the x values in this question.

Part (b)
In order to compute f'(x), you have to first deal with the absolute value function.  First, find the zero for this function.  It is obvious that the zero is x = 3/4.  When x < 3/4, f is negative.  When x > 3/4, f is positive.  That means that you can now define f(x) as a piecewise function using the squiggly bracket { like we see all the time in Lesson 3: Continuity:

f(x) =
  • -(4x - 3) when x < 3/4
  • 0 when x = 3/4
  • 4x - 3 when x > 3/4
Recall: In Assignment 2, question 9, we proved that an absolute value function is not differentiable at its zero, so f'(x) does not exist at x = 3/4.  (I think we may have just found a critical number!).  But, you can now compute and state f'(x) for x < 3/4 and x > 3/4
Question 4
Similar to my Lesson 9, question 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 cube root function is always defined since you can cube root both positive and negative numbers.

Personally, I would avoid product rule by multiplying the cube root of x 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, 8], one critical point and one singular point.
Question 5
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= -3, and so it is certainly continuous on [-2,1] and differentiable on (-2,1), therefore the Mean Value Theorem applies.

Hint: You should get two answers for c, but only one of them will be within [-2, 1].
Question 6
A classic curve sketch problem. Make sure you have done my Lesson 9, questions 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 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). 
  • If f(-x) is identical to f(x), you have proved it is even. 
  • If all you have to do is multiply the original f(x) by -1 to make it identical to f(-x), you have proven that f(-x) = -f(x), and, therefore, the function is odd.
  • Otherwise, it is neither.
Question 7
See my tips for question 6 above.  This one is quite similar to my Lecture Problem 4.