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Question: How do you evaluate the function \(f(x) = {e^x}\) at the value of \(x = 3.2\)?...

How do you evaluate the function f(x)=exf(x) = {e^x} at the value of x=3.2x = 3.2?

Explanation

Solution

This problem deals with finding the value of the function at a certain given value of the variable xx. Here the given function is an exponential function. An exponential function is defined as a function with a positive constant other than 1 raised to a variable exponent. A function is evaluated by solving at a specific input value. An exponent model can be found when the growth rate and initial value are known.

Complete step-by-step solution:
Given that the function is f(x)=exf(x) = {e^x}.
ex{e^x} is a transcendental function meaning that is both irrational and cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication and root extraction.
So here the function ex{e^x} cannot be expressed as a fraction, except in the trivial case at x=0x = 0, the root of any polynomial with rational coefficients or the sum of any finite series. Thus, it can only ever be approximated by a number of any base.
The definition of ex{e^x} is given below, which is given by:
ex=n=0xnn!\Rightarrow {e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}}, here this converges xR\forall x \in \mathbb{R}
Considering the second definition, from here we can approximate ex{e^x} at x=3.2x = 3.2, as given below:
ex=1+x1!+x22!+x33!+x44!+....\Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ....
At x=3.2x = 3.2, as shown below:
e3.2=1+3.2+(3.2)22!+(3.2)33!+(3.2)44!+....\Rightarrow {e^{3.2}} = 1 + 3.2 + \dfrac{{{{\left( {3.2} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {3.2} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {3.2} \right)}^4}}}{{4!}} + ....
e3.224.53253\Rightarrow {e^{3.2}} \approx 24.53253

The value of f(x)=exf(x) = {e^x} at x=3.2x = 3.2 is equal to 24.53253

Note: Please note that there are some basic rules that apply to exponential functions. The parental exponential function f(x)=bxf(x) = {b^x} always has a horizontal asymptote at y=0y = 0, except when b=1b = 1. You can’t raise a positive number to any power and get 0 or a negative number. You can’t multiply before you deal with the exponent.