Analyze and sketch a graph of the function. Find any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to
verify your results. (Order your answers from smallest to largest , then from smallest to largest . If an answer does not exist, enter ONE.)

intercept

relative minimum
relative maximum
(
point of inflection
(

Ask by Guerrero Ramos. in the United States

Mar 31,2025

Question

Analyze and sketch a graph of the function. Find any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to
verify your results. (Order your answers from smallest to largest , then from smallest to largest . If an answer does not exist, enter ONE.)

intercept

relative minimum
relative maximum
(
point of inflection
(

Ask by Guerrero Ramos. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

- **Y-intercept**: Set $$ x = 0 $$:
  $$
  f(0) = \frac{0^3}{0^2 - 36} = 0
  $$
  So, the y-intercept is $$ (0, 0) $$.

- **X-intercepts**: Set $$ f(x) = 0 $$:
  $$
  \frac{x^3}{x^2 - 36} = 0 \implies x^3 = 0 \implies x = 0
  $$
  So, the x-intercept is also $$ (0, 0) $$.

- **First Derivative**: 
  $$
  f'(x) = \frac{(3x^2)(x^2 - 36) - (x^3)(2x)}{(x^2 - 36)^2}
  $$
  Simplifying the numerator:
  $$
  3x^2(x^2 - 36) - 2x^4 = 3x^4 - 108x^2 - 2x^4 = x^4 - 108x^2
  $$
  Setting $$ f'(x) = 0 $$:
  $$
  x^4 - 108x^2 = 0 \implies x^2(x^2 - 108) = 0 \implies x = 0, \pm \sqrt{108} = \pm 6\sqrt{3}
  $$
  
- **Second Derivative Test**: We will evaluate $$ f''(x) $$ at these critical points to determine if they are minima or maxima.

- **Second Derivative**: 
  We will find $$ f''(x) $$ and set it to zero to find points of inflection.

- **Vertical Asymptotes**: Set the denominator to zero:
  $$
  x^2 - 36 = 0 \implies x = \pm 6
  $$
- **Horizontal Asymptote**: As $$ x \to \infty $$ or $$ x \to -\infty $$, the leading terms dominate:
  $$
  f(x) \approx \frac{x^3}{x^2} = x \implies \text{No horizontal asymptote.}
  $$
Now, let's calculate the second derivative and evaluate the critical points to find relative extrema and points of inflection.
**Intercepts:** - $$ (0, 0) $$ **Relative Extrema:** - **Minimum:** $$ (6\sqrt{3}, -108) $$ - **Maximum:** $$ (-6\sqrt{3}, 108) $$ **Points of Inflection:** - $$ (0, 0) $$ **Asymptotes:** - **Vertical Asymptotes:** $$ x = 6 $$ and $$ x = -6 $$ - **No Horizontal Asymptote**

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

Intercepts:

Relative Extrema:

Points of Inflection:

Asymptotes:

Summary of Results:

preguntas relacionadas

Latest Calculus Questions