Question

The range of the function $f(x)=\left(\frac{1}{3}\right)^{3+\sin x}$ is

• A. $\left[-\frac{1}{9},\frac{1}{81}\right]$
• B. $\left[-\frac{1}{9},\frac{1}{3}\right]$
• C. $\left[\frac{1}{9},\frac{1}{3}\right]$
• D. $\left[\frac{1}{81},\frac{1}{9}\right]$
• E. $\left[\frac{1}{81},\frac{1}{3}\right]$

Hint:

Logic Tip: Always pay attention to the base of an exponential function. If the base $b>1$, the function preserves inequality directions. If $0<b<1$ (like 1/3 here), the function reverses the inequality directions.

Answer 1

The Correct Option is D

Concept:
To find the range of an exponential function with a bounded exponent, first determine the minimum and maximum values (the bounds) of the exponent itself. Then, substitute these bounds into the exponential function to find the corresponding minimum and maximum outputs.
Step 1: Determine the range of the trigonometric component.
We know the standard range for the sine function for all real numbers $x$: $$-1 \le \sin x \le 1$$
Step 2: Determine the range of the entire exponent.
The exponent in the given function is $u = 3 + \sin x$. Add 3 to all parts of the sine inequality: $$3 - 1 \le 3 + \sin x \le 3 + 1$$ $$2 \le 3 + \sin x \le 4$$ Thus, the exponent ranges from $2$ to $4$.
Step 3: Apply the base to the exponent bounds.
The base of the exponential function is $\frac{1}{3}$. Since $\frac{1}{3}<1$, the function $y = (\frac{1}{3})^u$ is strictly decreasing. This means the *maximum* exponent will produce the *minimum* function value, and vice versa. Let's evaluate the function at the exponent boundaries: If the exponent is $4$: $f_{\text{min}} = \left(\frac{1}{3}\right)^4 = \frac{1}{81}$ If the exponent is $2$: $f_{\text{max}} = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$
Step 4: Write the final range interval.
The function's outputs continuously span from the minimum value to the maximum value: $$\text{Range} = \left[\frac{1}{81}, \frac{1}{9}\right]$$