Q. The inverse of the function (f(x)=2x+4) is (f^{-1}(x)=frac{x-4}{2}).

Q: How do you find the inverse of 2x+4?

Let y=2x+4. Solve: x=\frac{y-4}{2}. Swap variables to get the inverse: f^{-1}(x)=\frac{x-4}{2}.

Q: Is the inverse defined for all real numbers?

Yes. For f(x)=2x+4 the domain and range are all real numbers, so f^{-1}(x)=\frac{x-4}{2} is defined for every real x.

Q: Is f(x)=2x+4 invertible?

Yes. It is one-to-one because the slope 2 is nonzero, so every output has a unique input; therefore an inverse exists.

Q: How can I verify my inverse is correct?

Compose them: f(f^{-1}(x))=2\!\left(\frac{x-4}{2}\right)+4=x and f^{-1}(f(x))=\frac{2x+4-4}{2}=x. Both give x, so the inverse is correct.

Q: What does the graph of the inverse look like?

The inverse is the reflection of y=2x+4 across the line y=x. Its equation y=\frac{1}{2}x-2 has slope 1/2 and y-intercept -2.

Q: What about the inverse of a general linear function ax+b?

What about the inverse of a general linear function ax+b?

Q: Is the inverse the same as the reciprocal 1/(2x+4)?.

No. The inverse function is f^{-1}(x)=\frac{x-4}{2}. The reciprocal is \frac{1}{2x+4}; they are different operations and produce different expressions and values.

Q: How do I compute f^{-1}(6)?.

Evaluate f^{-1}(x)=\frac{x-4}{2} at x=6: f^{-1}(6)=\frac{6-4}{2}=1.

Q: How are the slopes of a linear function and its inverse related?

For f(x)=ax+b, the inverse has slope 1/a. Here a=2, so the inverse slope is 1/2.

Answer

Let \(y=2x+4\). Swap \(x\) and \(y\): \(x=2y+4\). Solve for \(y\): \(y=\frac{x-4}{2}\).
Inverse: \(f^{-1}(x)=\frac{x-4}{2}\).

Detailed Explanation

Find the inverse of the function

Given the function \( f(x) = 2x + 4 \).

Introduce a new variable for the output. Write the function as

\( y = 2x + 4 \).
Interchange the roles of the input and the output to obtain the equation that defines the inverse function. Replace y by x and x by y to get

\( x = 2y + 4 \).

Explanation: Exchanging x and y reflects across the line y = x, which yields the relation that the inverse must satisfy.
Solve this equation for y step by step.

First subtract 4 from both sides:

\( x – 4 = 2y \).

Explanation: Isolating the term containing y simplifies solving for y.

Next divide both sides by 2:

\( y = \dfrac{x – 4}{2} \).

Explanation: Dividing by the coefficient of y yields y explicitly in terms of x.
Rename y as the inverse function. Thus

\( f^{-1}(x) = \dfrac{x – 4}{2} \).
State domain and range.

Because \( f(x) = 2x + 4 \) is a linear function with domain all real numbers, the inverse also has domain all real numbers. Similarly, the range is all real numbers.
Verify by composition.

Compute \( f(f^{-1}(x)) \):

\( f\bigl(f^{-1}(x)\bigr) = 2\left(\dfrac{x – 4}{2}\right) + 4 = x – 4 + 4 = x \).

Compute \( f^{-1}(f(x)) \):

\( f^{-1}\bigl(f(x)\bigr) = \dfrac{(2x + 4) – 4}{2} = \dfrac{2x}{2} = x \).

Explanation: Both compositions return x, confirming the functions are inverses.

Final answer: \( f^{-1}(x) = \dfrac{x – 4}{2} \).

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Algebra FAQs

How do you find the inverse of \(2x+4\)?

Let \(y=2x+4\). Solve: \(x=\frac{y-4}{2}\). Swap variables to get the inverse: \(f^{-1}(x)=\frac{x-4}{2}\).

Is the inverse defined for all real numbers?

Yes. For \(f(x)=2x+4\) the domain and range are all real numbers, so \(f^{-1}(x)=\frac{x-4}{2}\) is defined for every real \(x\).

Is \(f(x)=2x+4\) invertible?

Yes. It is one-to-one because the slope \(2\) is nonzero, so every output has a unique input; therefore an inverse exists.

How can I verify my inverse is correct?

Compose them: \(f(f^{-1}(x))=2\!\left(\frac{x-4}{2}\right)+4=x\) and \(f^{-1}(f(x))=\frac{2x+4-4}{2}=x\). Both give \(x\), so the inverse is correct.

What does the graph of the inverse look like?

The inverse is the reflection of \(y=2x+4\) across the line \(y=x\). Its equation \(y=\frac{1}{2}x-2\) has slope \(1/2\) and \(y\)-intercept \(-2\).

What about the inverse of a general linear function \(ax+b\)?

Is the inverse the same as the reciprocal \(1/(2x+4)\)?.

No. The inverse function is \(f^{-1}(x)=\frac{x-4}{2}\). The reciprocal is \(\frac{1}{2x+4}\); they are different operations and produce different expressions and values.

How do I compute \(f^{-1}(6)\)?.

Evaluate \(f^{-1}(x)=\frac{x-4}{2}\) at \(x=6\): \(f^{-1}(6)=\frac{6-4}{2}=1\).

How are the slopes of a linear function and its inverse related?

For \(f(x)=ax+b\), the inverse has slope \(1/a\). Here \(a=2\), so the inverse slope is \(1/2\).

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