Let g(x)= arctan(x^3 + 3x^2 - 9x). Find where g(x) is increasing. I know the values of x which g(x) has a hortizontal tangent are x +-3 and x = 1, but I don't know of I should just graph it or use those to find where it increases.

Let g(x)= arctan(x^3 + 3x^2 - 9x). Find where g(x) is increasing. I know the values of x which g(x) has a hortizontal tangent are x +-3 and x = 1, but I don't know of I should just graph it or use those to find where it increases.

Mathematics For Machine Technology

8th Edition

ISBN:9781337798310

Author:Peterson, John.

Publisher:Peterson, John.

Chapter58: Achievement Review—section Five

Section: Chapter Questions

Problem 30AR: Determine dimension x to 3 decimal places.

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

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Expert Solution

Solution:

The function is $g (x) = a r c t a n (x^{3} + 3 x^{2} - 9 x)$

Obtain the derivative:

$\begin{array}{rcl} g' (x) & = & \frac{d}{d x} (\arctan (x^{3} + 3 x^{2} - 9 x)) \\ = & \frac{3 x^{2} + 6 x - 9}{{(x^{3} + 3 x^{2} - 9 x)}^{2} + 1} \end{array}$

Substitute $x = - 3$ : $\begin{array}{rcl} g' (- 3) & = & \frac{3 {(- 3)}^{2} + 6 (- 3) - 9}{{({(- 3)}^{3} + 3 {(- 3)}^{2} - 9 (- 3))}^{2} + 1} = 0 \end{array}$

Substitute $x = 1$ : $\begin{array}{rcl} g' (1) & = & \frac{3 {(1)}^{2} + 6 (1) - 9}{{({(1)}^{3} + 3 {(1)}^{2} - 9 (1))}^{2} + 1} = 0 \end{array}$

Hence, these values $x = - 3$ and $x = 1$ are critical values.

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