Compare the graphs of f(x)=x−−√+1 and g(x)=x−−√3+1 . Which one of the following options is true?

Question

(1 point)
Responses

Both functions are increasing on their domain.
Both functions are increasing on their domain.

Both functions have the same domain, x>0.
Both functions have the same domain, x greater than 0 .

Both functions have the same range, (−∞,∞).
Both functions have the same range, left parenthesis negative infinity comma infinity right parenthesis .

Both functions go through the points (−1,0) and (0,1).

Answer 1

To compare the graphs of the functions \( f(x) = \sqrt{x} + 1 \) and \( g(x) = \sqrt[3]{x} + 1 \), we can analyze their characteristics: domain, range, and monotonicity (increasing or decreasing nature).

1. Domain:

   • For \( f(x) = \sqrt{x} + 1 \), the square root function is defined only for \( x \geq 0 \). Therefore, the domain of \( f(x) \) is \( [0, \infty) \).
   • For \( g(x) = \sqrt[3]{x} + 1 \), the cube root function is defined for all real numbers. Therefore, the domain of \( g(x) \) is \( (-\infty, \infty) \).

2. Range:

   • The range of \( f(x) \) starts at \( 1 \) (when \( x = 0 \)) and goes to \( \infty \). Thus, the range of \( f(x) \) is \( [1, \infty) \).
   • The range of \( g(x) \) starts at \( 1 \) (when \( x = 0 \)) and goes to \( \infty \) as well. Additionally, as \( x \) goes to negative values, \( g(x) \) approaches values lower than \( 1 \). Thus, the range of \( g(x) \) is \( (-\infty, \infty) \).

3. Monotonicity:

   • Both \( f(x) = \sqrt{x} + 1 \) and \( g(x) = \sqrt[3]{x} + 1 \) are increasing functions. The derivative of \( f(x) \) is \( f'(x) = \frac{1}{2\sqrt{x}} \), which is positive for \( x > 0 \), hence \( f(x) \) is increasing on its domain \( [0, \infty) \). The derivative of \( g(x) \), \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} \), is also positive for all \( x ≠ 0 \), indicating that \( g(x) \) is increasing on its entire domain.

4. Specific Points:

   • Evaluating \( f(-1) \) is not applicable as -1 is not in the domain of \( f(x) \). For \( g(-1) \), we have \( g(-1) = \sqrt[3]{-1} + 1 = -1 + 1 = 0 \).
   • Evaluating \( f(0) \) gives \( f(0) = 0 + 1 = 1 \). For \( g(0) \), we have \( g(0) = \sqrt[3]{0} + 1 = 0 + 1 = 1 \).

Based on this comparison:

• Both functions are increasing on their domain. (This statement is true.)
• Both functions have the same domain, \( x > 0 \). (This statement is false; the domains are different.)
• Both functions have the same range, \( (-\infty, \infty) \). (This statement is false; \( f(x) \) has a range of \( [1, \infty) \) while \( g(x) \) has a range of \( (-\infty, \infty) \).)
• Both functions go through the points (-1, 0) and (0, 1). (This statement is false; \( f(-1) \) is not defined.)

The correct response is:

Both functions are increasing on their domain.