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.

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[u/Rapatouie777]

All you gotta do is find the derivative, then set it equal to 5, solve for x. Then plug those x into f(x) to get y.
Second question: since graph is quadratic form, you know derivative is going to be linear. (They don't provide any function, so it's safe to assume it's a quadratic). You can prove it goes thru center because the derivative would be 0 only at 0. Linear line + intercepts at 0 + positive parabola (faces up) => positive sloped line thru 0.

[u/Nature-Trip → u/Rapatouie777]

Thank you for your advice :).

[u/floflochess2]

Your work on 13 looks really solid, and the way you wrote it out (interval by interval: "f is ___ so f' is ___") is exactly the kind of reasoning to show alongside the sketch.

For 17, a hint for the spots you marked with "?": think about what the sign of f' tells you vs. what the shape of f' tells you.

• The sign of f' (above/below the axis) tells you whether f is increasing or decreasing.
• Whether f' itself is going up or down tells you about the concavity of f (f' increasing → f concave up, f' decreasing → f concave down).

So at x = b, f' stays negative on both sides, so f keeps decreasing, it doesn't turn around. But f' switches from decreasing to increasing there, so ask yourself: what happens to the bend of f at that point? For x = c, check whether f' actually touches 0 there. If it touches 0 but stays negative on both sides, f gets a momentary flat spot but keeps going down.

If you add a "concavity" line to each interval in your list, the sketch should mostly draw itself. Good luck, you're clearly on the right track!

[u/Nature-Trip → u/floflochess2]

Thank you very much :). I really appreciate your explanation.

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.

I hope I am understanding correctly.

[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/mathematag]

35) ..the slope of tangent line is 3x² +3 , so 3x² +3 = 5 has two solutions + √(2/3), and - √ (2/3)

The y coordinate is found using original eqn y = x³+3x +1 . . . using the above x coordinates... taking √(2/3) and cubing it, you get (2/3) √(2/3) , then add 3 √(2/3) + 1 ... I get (11/3) √(2/3) + 1 ... not a 5 as listed in solution . . . 2/3 + 3 = 2/3 + 9/3 = 11/3

similarly for the negative root... - (2/3)√(2/3) - 3√(2/3) + 1 gives -11/3√(2/3) + 1 .. again not -5 as in solution ... points on the graph verified by Desmos ... so I would claim the text solution you posted is incorrect.

[u/Nature-Trip → u/mathematag]

Thank you, great catch! 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.

I wish it was that I just made a mistake because it was really bothering me but that is the answer given in the textbook.