please check image calc 1 work only pls question 1 a b c d e Below is the transcription of an educational exercise on analyzing a cubic function.---1) **SHOW ALL WORK!** For the following function: $ f(x) = \frac{1}{3}x^3 - 3x^2 - 7x $ a) State the location of all vertical asymptotes and holes, if any: b) State all horizontal asymptotes, if any: c) State all local extrema (max and mins), if any, and the intervals of increasing/decreasing: d) State all points of inflection, if any, and the intervals of concave up/down: e) Sketch the graph of the function in the box to the right. Use a straight edge for your axes and indicate the scale (label the tick marks) on the axes. Include all points from above (and additional points, if necessary, to get an accurate sketch). *[There is a blank box provided for sketching]*2) In the space below, indicate the absolute extrema on the interval $[-3, 2]$ for $ f(x) = (x - 1)^{\frac{2}{3}} $. ---Note: The graph of the function should reflect the critical points, asymptotes, and behavior identified in parts (a) to (d).

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Description: please check image calc 1 work only pls question 1 a b c d e Below is the transcription of an educational exercise on analyzing a cubic function.---1) **SHOW ALL WORK!** For the following function: $ f(x) = \frac{1}{3}x^3 - 3x^2 - 7x $ a) State

Answered: 1 1) SHOW ALL WORK! For the following…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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calc 1 work only pls

question 1 a b c d e

Below is the transcription of an educational exercise on analyzing a cubic function.

---

1) **SHOW ALL WORK!** For the following function: $ f(x) = \frac{1}{3}x^3 - 3x^2 - 7x $

a) State the location of all vertical asymptotes and holes, if any:

b) State all horizontal asymptotes, if any:

c) State all local extrema (max and mins), if any, and the intervals of increasing/decreasing:

d) State all points of inflection, if any, and the intervals of concave up/down:

e) Sketch the graph of the function in the box to the right. Use a straight edge for your axes and indicate the scale (label the tick marks) on the axes. Include all points from above (and additional points, if necessary, to get an accurate sketch).

*[There is a blank box provided for sketching]*

2) In the space below, indicate the absolute extrema on the interval $[-3, 2]$ for $ f(x) = (x - 1)^{\frac{2}{3}} $.

---

Note: The graph of the function should reflect the critical points, asymptotes, and behavior identified in parts (a) to (d).

Expert Solution

Dear student,

Since you have posted multiple subparts, as per our policy we will answer first three subparts. Please repost the other subparts separately which you'd like to get answered.

Given,

$f (x) = \frac{1}{3} x^{3} - 3 x^{2} - 7 x$

On simplification, we get

$f (x) = \frac{x^{3} - 9 x^{2} - 21 x}{3}$

a).

Clearly denominator cannot be zero for any value of x.

There are no vertical asymptotes and holes.

b).

Since the degree of numerator (3) is greater than the degree of denominator (0), therefore there are no horizontal asymptotes.

Step 2

c).

Differentiating $f (x) = \frac{1}{3} x^{3} - 3 x^{2} - 7 x$ with respect to x, we get

$\begin{array}{rcl} f' (x) & = & \frac{1}{3} \cdot 3 x^{2} - 3 (2 x) - 7 (1) \\ = & x^{2} - 6 x - 7 \end{array}$

Now to find critical values, solve the first derivative by equating it to zero. That is,

$f' (x) = 0 \Rightarrow x^{2} - 6 x - 7 = 0 \Rightarrow (x - 7) (x + 1) = 0 \Rightarrow x = 7 o r x = - 1$

Therefore critical values are $x = 7 & x = - 1$ .

Now $f' (x) < 0$ to the left of $x = 7$ & $f' (x) > 0$ to the right of $x = 7$ .

And $f' (x) > 0$ to the left of $x = - 1$ & $f' (x) < 0$ to the right of $x = - 1$ .

Therefore $x = 7$ is the point of local minima and $x = - 1$ is the point of local maxima.

Now local minimum value is,

$\begin{array}{rcl} f (7) & = & \frac{1}{3} {(7)}^{3} - 3 {(7)}^{2} - 7 (7) \\ = & - \frac{245}{3} \end{array}$

& local maximum value is,

$\begin{array}{rcl} f (- 1) & = & \frac{1}{3} {(- 1)}^{3} - 3 {(- 1)}^{2} - 7 (- 1) \\ = & - \frac{1}{3} + 4 \\ = & \frac{11}{3} \end{array}$

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