Q.31 - 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.

$f\left(x\right)=\frac{2^x-4^x}{3^x}$

Step-by-Step Solution

Answer

The function has a real root at $x=0$.

The function has no discontinuity.

The function has no horizontal asymptote.

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.

$f\left(x\right)=\frac{2^x-4^x}{3^x}$

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:

$$\begin{gathered} f\left(x\right)=0 \\ \frac{2^x-4^x}{3^x}=0 \\ \frac{2^x\left(1-2^x\right)}{3^x}=0 \\ {\left(\frac{2}{3}\right)}^x\left(1-2^x\right)=0 \\ 1-2^x=0 \\ {2}^x=1 \\ x=0 \end{gathered}$$

Hence the root of the function is 0.

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.

Since $3^x$ cannot be negative, the function has no discontinuity.

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.

Here at both the positive and negative infinity, the function is in indeterminate form. Hence there are no horizontal asymptotes.

Step 5. Vertical asymptote of a function

The function has no discontinuity (undefined points). Hence there are no vertical asymptotes.

Step 6. Conclusion

The function has a real root at $x=0$.

The function has no discontinuity.

The function has no horizontal asymptote.

The function has no vertical asymptote.