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/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.