Calculus
Topics

# What are the absolute extrema of f(x)=x^3 - 3x + 1 in[0,3]?

May 25, 2016

#### Answer:

On $\left[0 , 3\right]$, the maximum is $19$ (at $x = 3$) and the minimum is $- 1$ (at $x = 1$).

#### Explanation:

To find the absolute extrema of a (continuous) function on a closed interval, we know that the extrema must occur at either crtical numers in the interval or at the endpoints of the interval.

$f \left(x\right) = {x}^{3} - 3 x + 1$ has derivative

$f ' \left(x\right) = 3 {x}^{2} - 3$.

$3 {x}^{2} - 3$ is never undefined and $3 {x}^{2} - 3 = 0$ at $x = \pm 1$.

Since $- 1$ is not in the interval $\left[0 , 3\right]$, we discard it.

The only critical number to consider is $1$.

$f \left(0\right) = 1$

$f \left(1\right) = - 1$ and

$f \left(3\right) = 19$.

So, the maximum is $19$ (at $x = 3$) and the minimum is $- 1$ (at $x = 1$).

##### Impact of this question
264 views around the world
You can reuse this answer
Creative Commons License
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9596685171127319, "perplexity": 703.3287973342447}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998288.34/warc/CC-MAIN-20190616162745-20190616184745-00005.warc.gz"}