Question

I am confused on this , I’ve tried twice and got it wrong.

Answer 1

Answer:

$f^-^1(x)=-x^2+6x-5 \ \text{for the domain}\ [3, \infty)$

Step-by-step explanation:

Consider the function  $f(x)=\sqrt{4-x}+3$ for the domain $(- \infty, 4]$.

Find $f^-^1(x)$, where f^(-1) is the inverse of f.

Also state the domain of f^(-1) in interval notation.

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We can start solving this problem by finding the inverse of f(x). This is done by switching the x- and y- variables, and solving for y.

• $y=\sqrt{4-x}+3 \rightarrow x=\sqrt{4-y}+3$
• $x=\sqrt{4-y}+3$

We can start solving for y by subtracting 3 from both sides of the equation.

• $x-3=\sqrt{4-y}$

Get rid of the radical by squaring both sides of the equation.

• $(x-3)^2=(\sqrt{4-y})^2$
• $(x-3)(x-3)=4-y$

Use FOIL to multiply the binomial (x-3) together.

• $x^2-3x-3x+9=4-y$

Combine like terms.

• $x^2-6x+9=4-y$

Subtract 4 from both sides of the equation.

• $x^2-6x+5=-y$

Divide both sides of the equation by -1.

• $-x^2+6x-5=y$
• $f^-^1(x)=-x^2+6x-5$

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The domain and range of a function are flipped for its inverse, meaning that to find the domain of the inverse function, you can find the range of the original function f(x), and that will be your inverse function's domain.

The range of $f(x)=\sqrt{4-x} +3$ is $y \geq 3$, since the vertical shift of the graph is at k = 3. You can also graph this function on a calculator to see that the graph does indeed start at y = 3.

Now that we know the domain and range of the original function, we know that these are flipped for the inverse function.

Original function:

• Domain: $x\leq 4$
• Range: $y\geq 3$

Inverse function:

• Domain: $x\geq 3$
• Range: $y\leq 4$

The final answer is:

The inverse $f^-^1(x)=-x^2+6x-5 \ \text{for the domain}\ [3, \infty)$.

You can also write the domain as: $x\geq 3$.