Question

To determine

(a) The domain and range of \(f(x) = \ln x + 2\).

(b) The \(x\)-intercept of the graph of \(f(x) = \ln x + 2\).

(c) The graph of \(f(x) = \ln x + 2\).

Short answer

(a) The domain of \(f(x) = \ln x + 2\) is \(x > 0\) and the range of \(f(x) = \ln x + 2\) is \(( - \infty ,\infty )\).

(b) The \(x\)-intercept of the graph of \(f(x) = \ln x + 2\) is \({e^{ - 2}}\).

(c) The \(x\)-intercept of the graph is \({e^{ - 2}} = 0.1353\).

Step-by-step solution

Given data

The given function is \(f(x) = \ln x + 2\).

Concept of Law of logarithm

Laws of logarithm:

Quotient law:\({\log _a}\left( {\frac{x}{y}} \right) = {\log _a}x - {\log _a}y\).

Product law:\({\log _a}(xy) = {\log _a}x + {\log _a}y\).

Power law:\({\log _a}\left( {{x^r}} \right) = r{\log _a}x\), where\(r\)is any real number.

Calculation of the domain and range of the function \(f(x) = \ln x + 2\)

(a)

The given function is \(f(x) = \ln x + 2\).

It is known that the \(\ln x\) is defined when \(x > 0\).

Thus, the domain of \(f(x) = \ln x + 2\) is \(x > 0\) as the number 2 does not make any change in the domain.

Also, it is known that the range of \(\ln x\) is \(( - \infty ,\infty )\).

Thus, the range of \(f(x) = \ln x + 2\) is \(( - \infty ,\infty )\) as 2 does not make any change in the range.

Calculation of the \(x\) – intercept of function \(f(x) = \ln x + 2\)

(b)

Set the given equation \(\ln x + 2 = 0\) and solve for \(x\) as shown below.

\(\begin{array}{c}\ln x = - 2\\{e^{\ln x}} = {e^{ - 2}}\\x = {e^{ - 2}}\end{array}\)

Thus, the \(x\)-intercept of \(f(x) = \ln x + 2\) is \({e^{ - 2}}\).

Plot of the graph of function \(f(x) = \ln x + 2\)

(c)

The graph of \(f(x) = \ln x + 2\) is shown below in Figure 1.

From Figure 1, it is noticed that the \(x\)-intercept of the graph is \({e^{ - 2}} = 0.1353\).