Question

How do you graph \[f\left( x \right)=2+\ln x\]?

Answer 1

Hint: This question is from the topic of exponential and logistic graphs which belongs to the chapter pre-calculus. In this question, we will draw the graph of \[f\left( x \right)=2+\ln x\]. In solving this question, we will first find the values of \[f\left( x \right)\] at x=0, 1, 2, 3, and \[\infty \]. After using these points and values of \[f\left( x \right)\] at those points, we will draw the graph of the given equation.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find the graph of the equation which is given in the question. The given equation is \[f\left( x \right)=2+\ln x\].
Let us first find out the value of \[f\left( x \right)\] at x=0. We can write
\[f\left( 0 \right)=2+\ln \left( 0 \right)\]
As we know that the value of \[\ln \left( 0 \right)\] is \[-\infty \], so we can write
\[\Rightarrow f\left( 0 \right)=2-\infty =-\infty \]
Now, let us first find out the value of \[f\left( x \right)\] at x=1. We can write
\[f\left( 1 \right)=2+\ln \left( 1 \right)\]
As we know that the value of \[\ln \left( 1 \right)\] is 0, so we can write
\[\Rightarrow f\left( 1 \right)=2+0=2\]
Now, let us find out the value of \[f\left( x \right)\] at x=2. We can write
\[f\left( 2 \right)=2+\ln \left( 2 \right)\]
As we know that the value of \[\ln \left( 2 \right)\] is 0.693, so we can write
\[\Rightarrow f\left( 2 \right)=2+0.693=2.693\]
Now, let us find out the value of \[f\left( x \right)\] at x=3.
\[f\left( 3 \right)=2+\ln \left( 3 \right)\]
Putting the value of \[\ln \left( 3 \right)\] as 1.0986, we can write
\[\Rightarrow f\left( 3 \right)=2+1.0986=3.0986\]
Now, let us find the value of \[f\left( x \right)\] at \[x=\infty \].
\[f\left( \infty \right)=2+\ln \left( \infty \right)\]
Putting the value of \[\ln \left( \infty \right)\] as \[\infty \] in the above equation, we can write
\[\Rightarrow f\left( \infty \right)=2+\infty =\infty \]
Now, we have got the following values:

\[x\]	0	1	2	3	\[\infty \]
\[f\left( x \right)\]	\[-\infty \]	2	2.693	3.0986	\[\infty \]

Now, we will use the above table and draw the graph of \[f\left( x \right)=2+\ln x\]

Now, we can see that at x=0, the value of f(x) is tending to negative or infinity at x equals to infinity, the value of f(x) is tending to infinity.


Note: We should have a better knowledge in the topic of exponential and logistic graphs to solve this type of question easily. We should remember the following values to solve this type of question easily:
Value of ‘e’ (that is exponential) = 2.71
\[\ln 0=-\infty \]
\[\ln 1=0\]
\[\ln 2=0.693\]
\[\ln 3=1.0986\]
\[\ln \infty =\infty \]
And, also remember that if \[\ln x=y\], then \[x={{e}^{y}}\]