
How do you evaluate the function $f\left( x \right)={{e}^{x}}$ at the value of x=3.2?
Answer
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Hint: The right approach for solving this question is the use of series and sequences. We solve the question by expressing ${{e}^{x}}$ as an infinite series. Now we must expand the series and then solve the series by substituting the value of 3.2 in place of x.
Complete step by step solution:
The number, e is a mathematical constant and irrational number which is numerically equal to 2.71828.
The expansion of the function is,
$e$ is a positive number so when raised to any power yields a positive value.
Thus, the domain of ${{e}^{x}}$ is a set of real numbers and the exponential function ranges from zero to infinity.
The derivative of the exponential function is always greater than zero. Hence the functions do not have any critical points which make it continuous throughout its domain.
The function ${{e}^{x}}$ can be expressed in series form as:
$\Rightarrow {{e}^{x}}=\displaystyle \lim_{n \to \infty }{{\left( \dfrac{1+x}{n} \right)}^{n}}$
The value of limit exists for every number that belongs to the set of real numbers or the domain of ${{e}^{x}}$
The sum of infinite terms of the exponential function which is equal to the above equation is given as,
$\Rightarrow {{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}}$
The irrational number e can be expressed as the sum of the above infinite series.
Now, the value of the function at x=3.2 can be evaluated by substituting the value of x in the following expression,
$\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}}$
Upon expanding it up to four or five terms, we get,
$\Rightarrow {{e}^{3.2}}=1+3.2+\dfrac{{{3.2}^{2}}}{2!}+\dfrac{{{3.2}^{3}}}{3!}+\dfrac{{{3.2}^{4}}}{4!}+........$
The value of the above series after calculations was found to be approximately equal to 24.53253
Hence, the function $f\left( x \right)={{e}^{x}}$ at the value of x=3.2 is approximately equal to 24.53253
Note: The exponential function of x can never be expressed as a rational number except for the value of x =0 and the sum of any finite series. We should remember the formulae listed above to solve the question easily otherwise it becomes a tedious job.
Complete step by step solution:
The number, e is a mathematical constant and irrational number which is numerically equal to 2.71828.
The expansion of the function is,
$e$ is a positive number so when raised to any power yields a positive value.
Thus, the domain of ${{e}^{x}}$ is a set of real numbers and the exponential function ranges from zero to infinity.
The derivative of the exponential function is always greater than zero. Hence the functions do not have any critical points which make it continuous throughout its domain.
The function ${{e}^{x}}$ can be expressed in series form as:
$\Rightarrow {{e}^{x}}=\displaystyle \lim_{n \to \infty }{{\left( \dfrac{1+x}{n} \right)}^{n}}$
The value of limit exists for every number that belongs to the set of real numbers or the domain of ${{e}^{x}}$
The sum of infinite terms of the exponential function which is equal to the above equation is given as,
$\Rightarrow {{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}}$
The irrational number e can be expressed as the sum of the above infinite series.
Now, the value of the function at x=3.2 can be evaluated by substituting the value of x in the following expression,
$\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}}$
Upon expanding it up to four or five terms, we get,
$\Rightarrow {{e}^{3.2}}=1+3.2+\dfrac{{{3.2}^{2}}}{2!}+\dfrac{{{3.2}^{3}}}{3!}+\dfrac{{{3.2}^{4}}}{4!}+........$
The value of the above series after calculations was found to be approximately equal to 24.53253
Hence, the function $f\left( x \right)={{e}^{x}}$ at the value of x=3.2 is approximately equal to 24.53253
Note: The exponential function of x can never be expressed as a rational number except for the value of x =0 and the sum of any finite series. We should remember the formulae listed above to solve the question easily otherwise it becomes a tedious job.
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