# Thread:

1. ## Finding the inverse of a function

Find the inverse of the function:

f(x) = (3/4) x^5 + 5

read "three-fourths x to the fifth plus 5"

Graphing my progress on my calculator as I go, the last point where I get an inverse is at

5^[sqrt] ((4/3)x - (20/3))
^(root with an index of 5)

If I am thinking correctly, you cannot have a root in the denominator, so I've been multiplying the whole thing above by

5^[sqrt](3^4)
^again, root with an index of 5

to get the fifth root out of the denominator, but this messes the whole equation up and it no longer mirrors the original function when I plug it into my graphing calculator. Is there some rule with roots about square roots that I am unaware of?

Help is much appreciated.

2. Originally Posted by happydino1
Find the inverse of the function:

f(x) = (3/4) x^5 + 5

read "three-fourths x to the fifth plus 5"

Graphing my progress on my calculator as I go, the last point where I get an inverse is at

5^[sqrt] ((4/3)x - (20/3))
^(root with an index of 5)

If I am thinking correctly, you cannot have a root in the denominator, so I've been multiplying the whole thing above by

5^[sqrt](3^4)
^again, root with an index of 5

to get the fifth root out of the denominator, but this messes the whole equation up and it no longer mirrors the original function when I plug it into my graphing calculator. Is there some rule with roots about square roots that I am unaware of?

Help is much appreciated.
Analytically to find the inverse of a function $y = f(x)$ we switch the roles of x and y to get $x = f(y)$, then solve for y: $y = g(x)$. g(x) is our inverse function.

So
$y = f(x) = \frac{3}{4}x^5 + 5$

Goes to
$x = \frac{3}{4}y^5 + 5$

$x - 5 = \frac{3}{4}y^5$

$y^5 = \frac{4}{3}(x - 5)$

$y = \sqrt[5]{\frac{4}{3}(x - 5)}$

-Dan

#### Search Tags

View Tag Cloud

Copyright © 2005-2013 Math Help Forum. All rights reserved.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8885561227798462, "perplexity_flag": "middle"}