
How do you evaluate the function $f(x) = {e^x}$ at the value of $x = 3.2$?
Answer
560.7k+ views
Hint: This problem deals with finding the value of the function at a certain given value of the variable $x$. 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 $f(x) = {e^x}$.
${e^x}$ 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 ${e^x}$ cannot be expressed as a fraction, except in the trivial case at $x = 0$, 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 ${e^x}$ is given below, which is given by:
$ \Rightarrow {e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} $, here this converges $\forall x \in \mathbb{R}$
Considering the second definition, from here we can approximate ${e^x}$ at $x = 3.2$, as given below:
$ \Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ....$
At $x = 3.2$, as shown below:
$ \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!}} + ....$
$ \Rightarrow {e^{3.2}} \approx 24.53253$
The value of $f(x) = {e^x}$ at $x = 3.2$ is equal to 24.53253
Note: Please note that there are some basic rules that apply to exponential functions. The parental exponential function $f(x) = {b^x}$ always has a horizontal asymptote at $y = 0$, except when $b = 1$. 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.
Complete step-by-step solution:
Given that the function is $f(x) = {e^x}$.
${e^x}$ 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 ${e^x}$ cannot be expressed as a fraction, except in the trivial case at $x = 0$, 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 ${e^x}$ is given below, which is given by:
$ \Rightarrow {e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} $, here this converges $\forall x \in \mathbb{R}$
Considering the second definition, from here we can approximate ${e^x}$ at $x = 3.2$, as given below:
$ \Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ....$
At $x = 3.2$, as shown below:
$ \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!}} + ....$
$ \Rightarrow {e^{3.2}} \approx 24.53253$
The value of $f(x) = {e^x}$ at $x = 3.2$ is equal to 24.53253
Note: Please note that there are some basic rules that apply to exponential functions. The parental exponential function $f(x) = {b^x}$ always has a horizontal asymptote at $y = 0$, except when $b = 1$. 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.
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