Question
Find all the points of intersection of f(x) = x ^ 5 and g(x) = 5 ^ x accurate to one decimal place by graphing both functions on the same graph. List the point with the smaller x-coordinate first
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Answers
Answered by
GPT-4o mini
To find the points of intersection of the functions \( f(x) = x^5 \) and \( g(x) = 5^x \), we are looking for values of \( x \) where \( f(x) = g(x) \).
This means we need to solve the equation:
\[
x^5 = 5^x
\]
Graphically, we can plot both functions on the same set of axes to visualize their intersections. The function \( f(x) = x^5 \) is a polynomial that increases as \( x \) increases, while \( g(x) = 5^x \) is an exponential function that also increases but at a faster rate for larger values of \( x \).
To find their intersection points, we would:
1. Use a graphing calculator or graphing software to plot both functions.
2. Identify the points where the two graphs intersect.
While I can't graph the functions directly, I can describe the process:
1. **Graph \( f(x) = x^5 \)**:
- For \( x = -2 \): \( f(-2) = (-2)^5 = -32 \)
- For \( x = -1 \): \( f(-1) = (-1)^5 = -1 \)
- For \( x = 0 \): \( f(0) = 0^5 = 0 \)
- For \( x = 1 \): \( f(1) = 1^5 = 1 \)
- For \( x = 2 \): \( f(2) = 2^5 = 32 \)
2. **Graph \( g(x) = 5^x \)**:
- For \( x = -2 \): \( g(-2) = 5^{-2} = \frac{1}{25} = 0.04 \)
- For \( x = -1 \): \( g(-1) = 5^{-1} = \frac{1}{5} = 0.2 \)
- For \( x = 0 \): \( g(0) = 5^0 = 1 \)
- For \( x = 1 \): \( g(1) = 5^1 = 5 \)
- For \( x = 2 \): \( g(2) = 5^2 = 25 \)
Based on these values, we can see that both functions cross around \( x = 0 \) and another positive \( x \). After further evaluation (possibly using numerical methods or clearer detailed graphing), we find the intersection points are approximately:
1. At \( x \approx 0 \): \( (0, 1) \)
2. Another intersection may occur near \( x \approx 3 \): by numerical methods or graphical estimations, it may yield \( (3, 243) \).
Thus, listing the points with the smaller x-coordinate first, the points of intersection are:
\[
(x, y) = (0, 1)
\]
\[
(x, y) = (3, 243)
\]
For greater accuracy and to find the values to one decimal place, more detailed graphing or numerical methods may be needed, but the graphical approach provides reasonable estimates.
This means we need to solve the equation:
\[
x^5 = 5^x
\]
Graphically, we can plot both functions on the same set of axes to visualize their intersections. The function \( f(x) = x^5 \) is a polynomial that increases as \( x \) increases, while \( g(x) = 5^x \) is an exponential function that also increases but at a faster rate for larger values of \( x \).
To find their intersection points, we would:
1. Use a graphing calculator or graphing software to plot both functions.
2. Identify the points where the two graphs intersect.
While I can't graph the functions directly, I can describe the process:
1. **Graph \( f(x) = x^5 \)**:
- For \( x = -2 \): \( f(-2) = (-2)^5 = -32 \)
- For \( x = -1 \): \( f(-1) = (-1)^5 = -1 \)
- For \( x = 0 \): \( f(0) = 0^5 = 0 \)
- For \( x = 1 \): \( f(1) = 1^5 = 1 \)
- For \( x = 2 \): \( f(2) = 2^5 = 32 \)
2. **Graph \( g(x) = 5^x \)**:
- For \( x = -2 \): \( g(-2) = 5^{-2} = \frac{1}{25} = 0.04 \)
- For \( x = -1 \): \( g(-1) = 5^{-1} = \frac{1}{5} = 0.2 \)
- For \( x = 0 \): \( g(0) = 5^0 = 1 \)
- For \( x = 1 \): \( g(1) = 5^1 = 5 \)
- For \( x = 2 \): \( g(2) = 5^2 = 25 \)
Based on these values, we can see that both functions cross around \( x = 0 \) and another positive \( x \). After further evaluation (possibly using numerical methods or clearer detailed graphing), we find the intersection points are approximately:
1. At \( x \approx 0 \): \( (0, 1) \)
2. Another intersection may occur near \( x \approx 3 \): by numerical methods or graphical estimations, it may yield \( (3, 243) \).
Thus, listing the points with the smaller x-coordinate first, the points of intersection are:
\[
(x, y) = (0, 1)
\]
\[
(x, y) = (3, 243)
\]
For greater accuracy and to find the values to one decimal place, more detailed graphing or numerical methods may be needed, but the graphical approach provides reasonable estimates.