The function $$ f(x)=x^{3}+4 $$ is one-to-one. (a) Find the inverse of $$ f $$ and check the answer. (b) Find the domain and the range of $$ f $$ and $$ f^{-1} $$ (c) Graph $$ f, f^{-1} $$, and $$ y=x $$ on the same coordinate axes. $$ \begin{array}{l}\text { (a) } f^{-1}(x)=\text { x } \ \text { (Simplify your answer. Use integers or fractions for any } \ \text { numbers in the expression.) }\end{array} $$

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UpStudy AI · Accepted Answer

We are given f(x) = x³ + 4. Since the cube function is one‐to‐one and adding 4 does not change that property, f is one‐to‐one and therefore has an inverse.

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Part (a). Find and Check the Inverse

1. Write the function with y:
  y = x³ + 4

2. Solve for x in terms of y:
  Subtract 4: y – 4 = x³
  Take the cube root: x = ∛(y – 4)

3. Swap the variables (replace y with x and x with y) to get the inverse:
  f⁻¹(x) = ∛(x – 4)

Check the answer:

• Compute f(f⁻¹(x)):
  f(f⁻¹(x)) = [ (∛(x – 4) )³] + 4 = (x – 4) + 4 = x

• Compute f⁻¹(f(x)):
  f⁻¹(f(x)) = ∛[(x³ + 4) – 4] = ∛(x³) = x

Since both compositions yield x, the inverse is verified.

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Part (b). Determine the Domain and Range

For the function f(x) = x³ + 4:

• Domain of f:
  All real numbers, written as (–∞, ∞).

• Range of f:
  Since x³ can produce any real number and adding 4 shifts the output, the range is also (–∞, ∞).

For the inverse f⁻¹(x) = ∛(x – 4):

Recall that the domain and range swap compared to f.

• Domain of f⁻¹:
  The domain of f⁻¹ is the same as the range of f, so it is (–∞, ∞).

• Range of f⁻¹:
  The range of f⁻¹ is the same as the domain of f, so it is (–∞, ∞).

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Part (c). Graphing f, f⁻¹, and y = x

To graph the three functions, keep the following in mind:

1. f(x) = x³ + 4:
  – This is a cubic function shifted upward by 4 units.
  – Typical shape: It increases from left to right with an inflection at the point where x³ “crosses” the origin (which here is shifted vertically).

2. f⁻¹(x) = ∛(x – 4):
  – This is the cube-root function shifted to the right by 4 units.
  – Its shape is the mirror image of f(x) with respect to the line y = x.

3. y = x:
  – This is the line at a 45° angle through the origin and serves as the “mirror line” between a function and its inverse.

Graphing Steps:
 a. Plot y = x as a dashed or distinct line.
 b. Plot f(x) = x³ + 4. For example, you could find sample points:
  • x = –3 → f(–3) = (–27) + 4 = –23
  • x = –1 → f(–1) = (–1) + 4 = 3
  • x = 0 → f(0) = 0 + 4 = 4
  • x = 1 → f(1) = 1 + 4 = 5
  • x = 2 → f(2) = 8 + 4 = 12
 c. Plot f⁻¹(x) = ∛(x – 4). Sample points can be chosen by picking x-values and computing:
  • x = –23 → f⁻¹(–23) = ∛(–23 – 4) = ∛(–27) = –3
  • x = 3 → f⁻¹(3) = ∛(3 – 4) = ∛(–1) = –1
  • x = 4 → f⁻¹(4) = ∛(0) = 0
  • x = 5 → f⁻¹(5) = ∛(5 – 4) = ∛(1) = 1
  • x = 12 → f⁻¹(12) = ∛(12 – 4) = ∛(8) = 2

Notice that the points on f and f⁻¹ are reflections of each other across the line y = x. For example, the point (0, 4) on f reflects to (4, 0) on f⁻¹.

Using these points, the graph can be sketched with all three functions on the same coordinate axes.

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Final Answers

(a) f⁻¹(x) = ∛(x – 4)

(b) Domain of f: (–∞, ∞); Range of f: (–∞, ∞)
      Domain of f⁻¹: (–∞, ∞); Range of f⁻¹: (–∞, ∞)

(c) The graph shows:
  • f(x) = x³ + 4, a cubic function shifted up 4 units.
  • f⁻¹(x) = ∛(x – 4), a cube-root function shifted right 4 units.
  • y = x, the line of reflection.
  When graphed, f and f⁻¹ are mirror images of each other with respect to the line y = x.
(a) $$ f^{-1}(x) = \sqrt[3]{x - 4} $$ (b) Domain and Range of $$ f $$ and $$ f^{-1} $$: - $$ f(x) = x^3 + 4 $$: - Domain: All real numbers ($$ -\infty, \infty $$) - Range: All real numbers ($$ -\infty, \infty $$) - $$ f^{-1}(x) = \sqrt[3]{x - 4} $$: - Domain: All real numbers ($$ -\infty, \infty $$) - Range: All real numbers ($$ -\infty, \infty $$) (c) The graph includes: - $$ f(x) = x^3 + 4 $$: A cubic function shifted upward by 4 units. - $$ f^{-1}(x) = \sqrt[3]{x - 4} $$: A cube-root function shifted to the right by 4 units. - $$ y = x $$: The line of reflection. - The functions $$ f $$ and $$ f^{-1} $$ are mirror images of each other across the line $$ y = x $$.

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