Question
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Let f(x)=(x−2)3+8.
Show that this function is one-to-one algebraically.
Find the inverse of f(x).
Let f(x)=(x−2)3+8.
Show that this function is one-to-one algebraically.
Find the inverse of f(x).
Answers
GPT-4o mini
To show that the function \( f(x) = (x - 2)^3 + 8 \) is one-to-one, we will demonstrate that it is either strictly increasing or strictly decreasing. A function is one-to-one if for any two distinct inputs \( a \) and \( b \), \( f(a) \neq f(b) \).
### Step 1: Show that the function is strictly increasing
To determine if the function is strictly increasing, we calculate its derivative:
\[
f'(x) = \frac{d}{dx}((x - 2)^3 + 8)
\]
Using the power rule:
\[
f'(x) = 3(x - 2)^2 \cdot (1) = 3(x - 2)^2
\]
### Step 2: Analyze the derivative
The derivative \( f'(x) = 3(x - 2)^2 \) is a square term multiplied by a positive constant \( 3 \). Therefore:
- \( (x - 2)^2 \geq 0 \) for all \( x \) (since a square is always non-negative).
- \( (x - 2)^2 = 0 \) when \( x = 2 \).
Thus, \( f'(x) \) is only equal to \( 0 \) at \( x = 2 \) and is positive for all other values of \( x \). Hence:
- \( f'(x) > 0 \) for all \( x \neq 2 \).
Since \( f'(x) > 0 \) for all \( x \neq 2 \), the function \( f(x) \) is strictly increasing everywhere except at the single point \( x = 2 \). This shows that \( f(x) \) is a one-to-one function.
### Step 3: Find the inverse of \( f(x) \)
To find the inverse function, we will solve the equation \( y = f(x) \) for \( x \):
\[
y = (x - 2)^3 + 8
\]
### Step 4: Rearrange the equation to solve for \( x \)
Subtract \( 8 \) from both sides:
\[
y - 8 = (x - 2)^3
\]
Now take the cube root of both sides:
\[
\sqrt[3]{y - 8} = x - 2
\]
Finally, add \( 2 \) to both sides to isolate \( x \):
\[
x = \sqrt[3]{y - 8} + 2
\]
### Step 5: Write the inverse function
We can denote the inverse function as \( f^{-1}(x) \):
\[
f^{-1}(x) = \sqrt[3]{x - 8} + 2
\]
### Conclusion
In summary, the function \( f(x) = (x - 2)^3 + 8 \) is one-to-one, and its inverse is given by:
\[
f^{-1}(x) = \sqrt[3]{x - 8} + 2
\]
### Step 1: Show that the function is strictly increasing
To determine if the function is strictly increasing, we calculate its derivative:
\[
f'(x) = \frac{d}{dx}((x - 2)^3 + 8)
\]
Using the power rule:
\[
f'(x) = 3(x - 2)^2 \cdot (1) = 3(x - 2)^2
\]
### Step 2: Analyze the derivative
The derivative \( f'(x) = 3(x - 2)^2 \) is a square term multiplied by a positive constant \( 3 \). Therefore:
- \( (x - 2)^2 \geq 0 \) for all \( x \) (since a square is always non-negative).
- \( (x - 2)^2 = 0 \) when \( x = 2 \).
Thus, \( f'(x) \) is only equal to \( 0 \) at \( x = 2 \) and is positive for all other values of \( x \). Hence:
- \( f'(x) > 0 \) for all \( x \neq 2 \).
Since \( f'(x) > 0 \) for all \( x \neq 2 \), the function \( f(x) \) is strictly increasing everywhere except at the single point \( x = 2 \). This shows that \( f(x) \) is a one-to-one function.
### Step 3: Find the inverse of \( f(x) \)
To find the inverse function, we will solve the equation \( y = f(x) \) for \( x \):
\[
y = (x - 2)^3 + 8
\]
### Step 4: Rearrange the equation to solve for \( x \)
Subtract \( 8 \) from both sides:
\[
y - 8 = (x - 2)^3
\]
Now take the cube root of both sides:
\[
\sqrt[3]{y - 8} = x - 2
\]
Finally, add \( 2 \) to both sides to isolate \( x \):
\[
x = \sqrt[3]{y - 8} + 2
\]
### Step 5: Write the inverse function
We can denote the inverse function as \( f^{-1}(x) \):
\[
f^{-1}(x) = \sqrt[3]{x - 8} + 2
\]
### Conclusion
In summary, the function \( f(x) = (x - 2)^3 + 8 \) is one-to-one, and its inverse is given by:
\[
f^{-1}(x) = \sqrt[3]{x - 8} + 2
\]