Q.32 - Calculus

Question

Find the roots, discontinuities, and horizontal and vertical asymptotes of the functions in Exercises 23–34. Support your answers by explicitly computing any relevant limits.

$f\left(x\right)=\frac{4^x-6\left(2^x\right)+5}{1-2^x}$

Step-by-Step Solution

Answer

The function has a real root at ${\log }_2\left(5\right)$.

The function has a discontinuity at 0.

The function has a horizontal asymptote $y=5$.

The function has no vertical asymptote.

Step 1. Given Information

Given to determine the roots, discontinuities, and horizontal and vertical asymptotes of the functions below.

$f\left(x\right)=\frac{4^x-6\left(2^x\right)+5}{1-2^x}$

Step 2. Roots of a function

The roots of a function are the values of x where the simplified function value is 0.

Simplifying the function and finding the root:

$$\begin{gathered} f\left(x\right)=0 \\ \frac{4^x-6\left(2^x\right)+5}{1-2^x}=0 \\ \frac{{\left(2^x\right)}^2-2^x-5\left(2^x\right)+5}{1-2^x}=0 \\ \frac{\left(-2^x\right)\left(-2^x+1\right)+5\left(-2^x+1\right)}{\left(1-2^x\right)}=0 \\ \frac{\left(1-2^x\right)\left(5-2^x\right)}{\left(1-2^x\right)}=0 \\ 5-2^x=0 \\ {2}^x=5 \\ x={\log }_2\left(5\right) \end{gathered}$$

Hence the root of the function is ${\log }_2\left(5\right)$

Step 3. Discontinuities of a function

A function has a discontinuity at a point $x=a$ where $f\left(a\right)\ne \underset{x\to a}{\lim }f\left(x\right)$

The function is undefined when the denominator is 0.

Here,

$$\begin{gathered} 1-2^x=0 \\ {2}^x=1 \\ x=0 \end{gathered}$$

Hence the function has a discontinuity at 0.

Step 4. Horizontal asymptote of a function

Horizontal asymptote of a function can be determined by the tangents at the infinity i.e. when x is positive or negative infinity.

When $x\to -\infty$

$$\begin{gathered} f\left(x\right)=\underset{x\to -\infty }{\lim }\frac{4^x-6\left(2^x\right)+5}{1-2^x} \\ f\left(x\right)=\frac{4^{-\infty }-6\left(2^{-\infty }\right)+5}{1-2^{-\infty }} \\ f\left(x\right)=\frac{0-6\left(0\right)+5}{1-0} \\ f\left(x\right)=5 \end{gathered}$$

Hence one horizontal asymptote is $y=5$

When $x\to \infty$

$$\begin{gathered} f\left(x\right)=\underset{x\to \infty }{\lim }\frac{4^x-6\left(2^x\right)+5}{1-2^x} \\ f\left(x\right)=\frac{4^{\infty }-6\left(2^{\infty }\right)+5}{1-2^{\infty }} \\ f\left(x\right)=\frac{\infty -6\left(\infty \right)+5}{1-\infty } \\ f\left(x\right)=\frac{\infty }{\infty } \end{gathered}$$

Here the asymptote is not defined.

Step 5. Vertical asymptote of a function

The vertical asymptote of a rational function are the zeroes of the denominator of the simplified function.

Here the simplified function is:

$f\left(x\right)=5-2^x$

Since there is not denominator, there is no vertical asymptote.

Step 6. Conclusion

The function has a real root at ${\log }_2\left(5\right)$

The function has a discontinuity at 0.

The function has a horizontal asymptote at $y=5$.

The function has no vertical asymptote.