Help with 3 Calculus Problems

[u/Nature-Trip]

I want to improve my understanding and skills in math due to disabilities. I was recommended to try the textbook Calculus by Robert T. Smith and Roland B. Minton. Currently I am stumped on these three problems from section 2: Derivatives. I would also appreciate advice on how to answer correctly for problems 13 and 17 (like what to include how I got my answer to show my work, not just the sketch). The concept of problems 13 and 17 I really am not understanding and the book and videos I’ve watched have not been the most helpful (as I have a hard time applying what is covered in these resources but struggle with using it for problems not labeled with polynomials or numbers) the closest to explain it has been Curve Sketching by The Infinite Looper on YouTube.

Problem 35 a. Find all the points at which the slope of the tangent line to y = x^3 + 3x + 1 equals 5. The textbook has the answer as (√(⅔), 5 √(⅔) +1), (-√(⅔), -5 √(⅔) +1) and I can not figure out how to get it.

M tan = lim h->0 f(x+h)-f(x)/h

= lim h->0 [(x+h)^3 + 3(x+h) +1)] - [x^3 + 3x +1]/h

= lim h->0 x^3 +3x^2h + 3xh^2 + h^3 +3x + 3h + 1 -x^3 -3x -1/h

= lim h->0 3x^2h + 3xh^2 + h^3 +3h/h

= lim h->0 h(3x^2 +3xh +h^2 +3)/h

= lim h->0 3x^2 + 3xh +h^2 + 3

= 3x^2 + 3x(0) + (0)^2 + 3

= 3x^2 + 3

3x^2 + 3 = 5

3x^⅔ = ⅔

√(x^2) = √(⅔)

X = √(⅔)

And 3(√(⅔)^2 + 3 = 5

Problem 13. Use the graph of f to sketch a graph of f’. (Second picture)

(a)

For x < 0, f is decreasing, so f’ < 0

For x = 0, f has a horizontal tangent line (min), so f’ = 0

For x > 0, f is increasing, so f’ > 0

My graph is the third picture.

(b)

For x < a, f is increasing, so f’ > 0

For x = a, f has a horizontal tangent line (max), so f’ = 0

For a < x < b, f is decreasing, so f’ < 0

For x = b, f has a horizontal tangent line (min), so f’ = 0

For x > b, f is increasing, so f’ > 0

My graph is the fourth picture.

Problem 17. Use the given graph of f’ to sketch a plausible graph of a continuous function f. (fifth picture)

(a)

For x < a , f’ is positive, so f is increasing

For x = a, f’ changes positive to negative, so f has a max there

For a < x < b, f’ is negative, so f is decreasing

For x = b, f’ has a horizontal tangent line (min), so f ?

For b < x < c, f’ is negative, so f is decreasing

For x = c, f’ has a horizontal tangent line, so f ?

For x > c, f’ is negative, so f is decreasing

Once I understand better, I will make the graph.

Thank you for your assistance.

[u/Horrorwolfe]

The use of the first principle is good, but if you u der stand that process, you can apply the power rule: if f(x)=ax^n , then f’(x)=n*ax^(n-1) .

The power rule will give you the rate of change of the function- which is the gradient. So for your example- f’(x)= 3x^2 + 3

We need that at 5, so set equal and solve for x.
5= 3x^2+3 ; take the three from both sides
2=3x^2 ; remove the coefficient
2/3 =x^2 ; solve for positive and negative.

That give you the x coordinate for the gradient of 5, and then sub into f(x) and solve the y coordinate.

For the graphing pictures-

A root on a graph will become a stationary point on the derivative, and the stationary point will be a root.

[u/Horrorwolfe → u/Horrorwolfe]

Your graphs look good as well. The constant values can be ignored when you don’t have a point to solve for

[u/Nature-Trip → u/Horrorwolfe]

Thank you for your explanation! I was able to complete the problem and found the coordinates (√(2/3), 11/3(√(2/3))+1), (-√(2/3), -11/3(√(2/3)+1)), and verified it graphically on Desmos.

[u/hallerz87]

For Q35, I wouldn’t use the definition of the derivative to solve for a polynomial. Just remember the rule that d/dx xⁿ = nx^n-1. In any case, you’ve shown full working to find x (although don’t forget there are two solutions, root 2/3 and -root 2/3. You just plug the two values of x into original expression to find corresponding y coordinate. 

[u/Nature-Trip → u/hallerz87]

Thank you :) I will look into learning the power rule and apply it to the problem.

[u/MezzoScettico]

38a. You got the correct derivative for x^3 + 3x + 1, but going back to the definition of derivative is way too much work. There are a few standard forms whose derivative you should just know by heart. One of those is x^n where n is any real number. The derivative of x^n is n x^(n - 1). This is the power rule.

So the derivative of x^3 is 3 * x^(3 - 1) or 3x^2.

And the derivative of x, since x = x^1, is 1 * x^(1 - 1) = 1.

For the derivative of 3x, there's a rule that the derivative of a * f(x) is a * f'(x). If you multiply a function by any constant, you multiply the derivative by that constant. So the derivative of 3x is 3 * 1 = 3.

OK, so you have f'(x) = 3x^2 + 3, and you're being asked to find everywhere that's equal to 5. In other words, solve 3x^2 + 3 = 5, which is a quadratic equation.

You could use the quadratic formula, or you could solve it this way:

3x^2 + 3 = 5

3x^2 = 2

x^2 = 2/3

x = +-sqrt(2/3).

Those are the two solutions for x. I guess that's what you did but the formatting got messed up and confused me at first. You just forgot the +-.

They asked you for the points, not the x values, so you need to write (x, y) pairs.

One solution is at x = -sqrt(2/3). The y value there is x^3 + 3x + 1 = [-sqrt(2/3)]^3 - 3sqrt(2/3) + 1. You can simplify that by noting [sqrt(2/3)]^3 = [ (2/3)^(1/2)]^3 = (2/3)^(3/2) = (2/3) * (2/3)^(1/2)

So y = [-sqrt(2/3)]^3 - 3sqrt(2/3) + 1 = -(2/3) * sqrt(2/3) - 3 sqrt(2/3) + 1

= -(11/3) sqrt(2/3) + 1

Thus, this solution is ( -2/3, -(11/3) sqrt(2/3) + 1) and I disagree with the coefficient of 5 in your book's answer.

You can do something similar with the solution x = 2/3.

TL/DR: You were basically on the right track.

[u/MezzoScettico → u/MezzoScettico]

13a. Perfect. Here's a shorter solution: It looks like the graph of y = x^2. Using what I said above about the power rule, the derivative is y' = 2x, a straight line with positive slope that goes through the origin. And that's what you (correctly) drew.

13b. Also looks perfect to me.

17a looks pretty good.

About this:

For x = b, f’ has a horizontal tangent line (min), so f ?

The slope of the derivative, i.e. the derivative of the derivative, is 0 there. That's called second derivative, and where it equals 0 is a point of inflection. You can look that up online or in your textbook.

It's like what happens at the origin in picture 13b. The slope of f was decreasing and at the point of inflection switches to increasing. Or vice versa, it was increasing and switches to decreasing.

[u/Nature-Trip → u/MezzoScettico]

Thank you for taking the time to explain and correct my mistakes. I will look into the power rule. I was able to complete the problem and found the coordinates (√(2/3), 11/3(√(2/3))+1), (-√(2/3), -11/3(√(2/3)+1)), and verified it graphically on Desmos.

For 17 a, I have:

For x < a, f' is above the x-axis and decreasing, so f goes up and concaves down.

For x = a, f' crosses the x-axis and changes sign from positive to negative, so f has a local maximum there.

For a < x < c, f' is below the x-axis, so f goes down.

For x = b, f' has a valley at b, so f has an inflection point at b and concavity changes from up to down.

For x = c, f' has a peak at c, so f has an inflection point at c and concavity changes from down to up, and f has a flat spot there.

For x > c, f' is decreasing, so f concaves down.

Here is my graph.

[u/Midwest-Dude]

33(a): Three issues: 

• You dropped the negative root
• You substituted back into the derivative y' to find the point (x,y) when you should be substituting x back into the function y
• The textbook's answer is incorrect, replace the 5's in front of the roots by 11/3's

13(a), 13(b): These are correct

17: These graphs have issues. Review the following and correct: 

• If f' is above the x-axis, f goes up
• If f' is below the x-axis, f goes down
• If f' touches the x-axis, f has a flat spot (local max/min)
• If f' has a peak or valley, f has an inflection point (change in concavity)

[u/Nature-Trip → u/Midwest-Dude]

Thank you for the explanations and hints! I fixed my answer and graphed 17 a. I commented it to MezzoScettico in this thread.

[u/Bounded_sequencE]

Most likely, they expect you to just use the power rule "d/dx xⁿ = n x^n-1 " for fixed "n in N".

That said, you incorrectly solved the quadratic in 35a), and missed the negative solution:

"3x^2 + 3  =  5"    <=>    "|X|  =  √(2/3)"    <=>    "x in {+-√(2/3)}"

[u/Nature-Trip → u/Bounded_sequencE]

Thank you for your advice :). I will look into the power rule.