Why is a rational function, not a transcendental function?

Question:

Why is a rational function, not a transcendental function?

Rational Functions

A function with one independent variable with form {eq}\displaystyle \ f(x) \, = \, \frac{ \ g(x) }{ h(x) } {/eq} is called rational function, where {eq}\displaystyle \ g(x) \; \text{ & } \; \ h(x) {/eq} are polynomial functions, and {eq}\displaystyle \ h(x) \, \neq \, 0 {/eq}

So, a rational function is different from a transcendental function, we are going to identify some calculus rules that determine the reasons for their differences.

Answer and Explanation:

A rational function is a function with the form:

{eq}\displaystyle \ f(x) \, = \, \frac{ g(x) }{ h(x) } \; \text{ with } \; h(x) \neq 0 {/eq} where {eq}\displaystyle g(x) \; \text{ & } \; h(x) {/eq} are polynomial functions.

Also, a polynomial function has the form:

{eq}\displaystyle \ f(x) \, = \, a_n x^n+a_{n-1}x^{n-1}+ \; \text{ ... } \; a_1x+a_0 {/eq}

where {eq}\displaystyle n {/eq} is a positive integer, and {eq}\displaystyle a_n \neq 0 {/eq}

Furthermore:

A transcendental function is any exponent, logarithm or trigonometric function.

An exponent function is a function with the form:

{eq}\displaystyle \ f(x) \, = \, ax^{bx} {/eq}

A logarithm function is a function with the form:

{eq}\displaystyle \ f(x) \, = \, a\log(bx) {/eq}

A few trigonometric functions are:

{eq}\displaystyle y \, = \, \sin(x) \\ \displaystyle y \, = \, \cos(x) \\ \displaystyle y \, = \, \frac{ \sin(x) }{ \cos(x) } \\ \displaystyle y= \frac{ 1 }{ \sin(x) } \\ \displaystyle y= \frac{ 1 }{ \cos(x) } \\ \displaystyle y= \frac{ 1 }{ \tan(x) } \\ {/eq}


We can conclude, a rational function can't be a trascendental function.


Learn more about this topic:

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Rational Function: Definition, Equation & Examples

from GMAT Prep: Help and Review

Chapter 10 / Lesson 11
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