Asked by shmunklee
an exponential function is described by the equation h (x) = 49(2/7)^x. which inequality describes the range of this function
a. -∞ < y < ∞
b. o < y < 16
c. y >= 0
d. y > 0
a. -∞ < y < ∞
b. o < y < 16
c. y >= 0
d. y > 0
Answers
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Answered by
GPT-4o mini
To determine the range of the exponential function given by the equation \( h(x) = 49\left(\frac{2}{7}\right)^x \), we need to consider the behavior of the function as \( x \) varies.
1. **Identify the base of the exponential function**: The base here is \( \frac{2}{7} \), which is between 0 and 1. As \( x \) increases, \( \left(\frac{2}{7}\right)^x \) approaches 0 but never actually reaches 0.
2. **Evaluate the function at specific points**:
- For \( x = 0 \):
\[
h(0) = 49\left(\frac{2}{7}\right)^0 = 49(1) = 49
\]
- As \( x \) approaches \( \infty \):
\[
h(x) \to 49(0) = 0
\]
3. **Range of the function**: Since \( h(x) \) decreases from \( 49 \) to \( 0 \) but never actually reaches \( 0 \), the function can take any positive value but cannot equal 0.
Given this analysis, the correct inequality describing the range of the function is:
**d. \( y > 0 \)**.
This means the output of the function is always positive.
1. **Identify the base of the exponential function**: The base here is \( \frac{2}{7} \), which is between 0 and 1. As \( x \) increases, \( \left(\frac{2}{7}\right)^x \) approaches 0 but never actually reaches 0.
2. **Evaluate the function at specific points**:
- For \( x = 0 \):
\[
h(0) = 49\left(\frac{2}{7}\right)^0 = 49(1) = 49
\]
- As \( x \) approaches \( \infty \):
\[
h(x) \to 49(0) = 0
\]
3. **Range of the function**: Since \( h(x) \) decreases from \( 49 \) to \( 0 \) but never actually reaches \( 0 \), the function can take any positive value but cannot equal 0.
Given this analysis, the correct inequality describing the range of the function is:
**d. \( y > 0 \)**.
This means the output of the function is always positive.
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