Question

If the derivative of a continuous function does not change its sign on its entire domain, then the function will be one-one. Analyze the function $f(x) = 1+\sqrt{x}$ with domain $[0,\infty)$ and codomain $[0,\infty)$.

• A. one-one and onto
• B. one-one but not onto
• C. onto but not one-one
• D. neither one-one nor onto

Answer 1

Answer: (B)

one-one but not onto

Answer 2

The statement provided is correct: a continuous function whose derivative does not change sign on its domain is one-to-one.

For the function $f(x) = 1 + \sqrt{x}$:

1. Continuity: $f(x)$ is continuous on its domain $[0, \infty)$.
2. Derivative: $f'(x) = \frac{1}{2\sqrt{x}}$. For $x > 0$, $f'(x) > 0$. The derivative is always positive on $(0, \infty)$, so it does not change sign. This confirms the function is one-to-one.
3. One-to-One Check: If $f(x_1) = f(x_2)$, then $1 + \sqrt{x_1} = 1 + \sqrt{x_2}$, which implies $\sqrt{x_1} = \sqrt{x_2}$, and thus $x_1 = x_2$. The function is indeed one-to-one.
4. Onto Check: The domain is $[0, \infty)$. The range of $\sqrt{x}$ for $x \in [0, \infty)$ is $[0, \infty)$. Therefore, the range of $f(x) = 1 + \sqrt{x}$ is $[1, \infty)$. The codomain is given as $[0, \infty)$. Since the range $[1, \infty)$ is not equal to the codomain $[0, \infty)$, the function is not onto.

Thus, the function is one-to-one but not onto.