Questions: Topic 9: Graphing Polynomial Functions Solve the problem. f(x) = √(x + 3) a. Graph f(x) b. Write an equation for f⁻¹(x) c. Write the domain of f⁻¹ in interval notation

Graph \( f(x) = \sqrt{x+3} \)

Identify the domain of \( f(x) \)

The function \( f(x) = \sqrt{x+3} \) is defined when the expression under the square root is non-negative. Therefore, \( x+3 \geq 0 \), which implies \( x \geq -3 \).

Plot the function

The function \( f(x) = \sqrt{x+3} \) is a transformation of the basic square root function, shifted 3 units to the left. The graph starts at \( x = -3 \) and increases as \( x \) increases.

\(\boxed{\text{Graph of } f(x) = \sqrt{x+3} \text{ is plotted.}}\)

Write an equation for \( f^{-1}(x) \)

Express \( y = f(x) \) in terms of \( x \)

Start with \( y = \sqrt{x+3} \).

Solve for \( x \) in terms of \( y \)

Square both sides to get \( y^2 = x+3 \). Then, solve for \( x \) to get \( x = y^2 - 3 \).

Express the inverse function

The inverse function is \( f^{-1}(x) = x^2 - 3 \).

\(\boxed{f^{-1}(x) = x^2 - 3}\)

Write the domain of \( f^{-1} \) in interval notation

Determine the domain of \( f^{-1}(x) \)

Since \( f(x) = \sqrt{x+3} \) has a range of \([0, \infty)\), the domain of \( f^{-1}(x) \) is \([0, \infty)\).

\(\boxed{[0, \infty)}\)

\(\boxed{\text{Graph of } f(x) = \sqrt{x+3} \text{ is plotted.}}\) \(\boxed{f^{-1}(x) = x^2 - 3}\) \(\boxed{[0, \infty)}\)

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