Find the range of real-valued exponential functions using different techniques such as inverse functions and algebraic analysis.
Examples with Solutions
Example 1
Find the range of function \( f \) defined by:
\[ f(x) = e^{-x+2} \]View Solution
Let us first write the function as an equation:
\[ y = e^{-x+2} \]Solve the equation for \( x \):
\[ -x + 2 = \ln(y) \] \[ x = 2 - \ln(y) \]\( x \) is a real number if \( y > 0 \) (since the argument of \( \ln y \) must be positive). Hence, the range of function \( f \) is given by:
\[ y > 0 \quad \text{or in interval form} \quad (0, +\infty) \]The graph of \( f \) below illustrates the range graphically:
Example 2
Find the range of function \( f \) defined by:
\[ f(x) = e^{2x+1} + 3 \]View Solution
Write the given function as an equation:
\[ y = e^{2x+1} + 3 \]Solve the equation for \( x \):
\[ y - 3 = e^{2x+1} \] \[ 2x + 1 = \ln(y - 3) \] \[ x = \frac{1}{2}(\ln(y - 3) - 1) \]\( x \) is a real number for \( y - 3 > 0 \) (the argument of \( \ln(y - 3) \) must be positive). The range of the given function is then given by:
\[ y > 3 \quad \text{or in interval form} \quad (3, +\infty) \]The graph of \( f \) below illustrates the range graphically:
Example 3
Find the range of function \( f \) defined by:
\[ f(x) = e^{x^2} + 1 \]View Solution
Write the given function as an equation:
\[ y = e^{x^2} + 1 \]Solve the equation for \( x \):
\[ y - 1 = e^{x^2} \] \[ x^2 = \ln(y - 1) \] \[ x = \pm\sqrt{\ln(y - 1)} \]The above solutions are real if:
\[ \ln(y - 1) \geq 0 \] \[ y - 1 \geq 1 \implies y \geq 2 \]Hence, the range of the given function is given by:
\[ y \geq 2 \quad \text{or in interval form} \quad [2, +\infty) \]The graph of \( f \) below illustrates the range graphically:
Example 4
Find the range of function \( f \) defined by:
\[ f(x) = -2e^{-x^2} + 3 \]View Solution
Write the given function as an equation:
\[ y = -2e^{-x^2} + 3 \]Solve for \( x \):
\[ y - 3 = -2e^{-x^2} \] \[ e^{-x^2} = \frac{y - 3}{-2} \] \[ -x^2 = \ln\left[\frac{y - 3}{-2}\right] \] \[ x = \pm\sqrt{-\ln\left[\frac{y - 3}{-2}\right]} \]\( x \) is real if the argument of \( \ln \) is positive and the radicand is positive or zero:
\[ \frac{y - 3}{-2} > 0 \quad \text{and} \quad -\ln\left[\frac{y - 3}{-2}\right] \geq 0 \]The solution set of \(\frac{y - 3}{-2} > 0\) is \( y < 3 \).
The solution set of \(-\ln\left[\frac{y - 3}{-2}\right] \geq 0\) is given by \(\frac{y - 3}{-2} \leq 1\), which yields \( y \geq 1 \).
Combining these conditions, the range of \( f \) is given by:
\[ 1 \leq y < 3 \quad \text{or in interval form} \quad [1, 3) \]The graph of \( f \) below illustrates the range graphically:
