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Question: If the derivative of a continuous function does not change its sign on its entire domain, then the f...

If the derivative of a continuous function does not change its sign on its entire domain, then the function will be one-one.

Answer

The statement is correct. A continuous function whose derivative does not change sign on its domain is strictly monotonic, and therefore one-to-one. For the given examples:

  1. f(x)=1+xf(x)=1+\sqrt{x} on [0,∞)[0,\infty) has f′(x)=12x>0f'(x) = \frac{1}{2\sqrt{x}} > 0 for x>0x>0. It is one-to-one but not onto as its range is [1,∞)[1,\infty) while codomain is [0,∞)[0,\infty).
  2. f(x)=2x3−3x2+4x−1f(x)=2x^3-3x^2+4x-1 on RR has f′(x)=6x2−6x+4f'(x) = 6x^2-6x+4. The discriminant is (−6)2−4(6)(4)=36−96=−60<0(-6)^2 - 4(6)(4) = 36-96 = -60 < 0. Since the leading coefficient is positive, f′(x)>0f'(x) > 0 for all x∈Rx \in R. It is strictly increasing and thus one-to-one. As a cubic polynomial, its range is RR, so it is also onto.
Explanation

Solution

The core principle is that if a continuous function's derivative maintains a consistent sign (always positive or always negative) across its entire domain, the function is strictly monotonic (either strictly increasing or strictly decreasing). Strictly monotonic functions are inherently one-to-one because each output value corresponds to a unique input value.

Analysis of Example 1: f(x)=1+xf(x) = 1 + \sqrt{x} on the domain [0,∞)[0, \infty).

  • Continuity: The function is continuous on [0,∞)[0, \infty).
  • Derivative: For x>0x > 0, f′(x)=12xf'(x) = \frac{1}{2\sqrt{x}}.
  • Sign of Derivative: For all x∈(0,∞)x \in (0, \infty), f′(x)>0f'(x) > 0. The derivative does not change sign.
  • One-to-One: Since f′(x)>0f'(x) > 0, the function is strictly increasing and thus one-to-one.
  • Onto: The range of f(x)f(x) for x∈[0,∞)x \in [0, \infty) is [1,∞)[1, \infty). The given codomain is [0,∞)[0, \infty). Since the range is not equal to the codomain, the function is not onto.
  • Conclusion for Example 1: One-to-one but not onto.

Analysis of Example 2: f(x)=2x3−3x2+4x−1f(x) = 2x^3 - 3x^2 + 4x - 1 on the domain RR.

  • Continuity: This is a polynomial function, so it is continuous on RR.
  • Derivative: f′(x)=6x2−6x+4f'(x) = 6x^2 - 6x + 4.
  • Sign of Derivative: To determine the sign of f′(x)f'(x), we examine its discriminant. For the quadratic ax2+bx+cax^2+bx+c, the discriminant is Δ=b2−4ac\Delta = b^2 - 4ac. Here, a=6a=6, b=−6b=-6, c=4c=4. Δ=(−6)2−4(6)(4)=36−96=−60\Delta = (-6)^2 - 4(6)(4) = 36 - 96 = -60. Since Δ<0\Delta < 0 and the leading coefficient (a=6a=6) is positive, the quadratic f′(x)f'(x) is always positive for all x∈Rx \in R. The derivative does not change sign.
  • One-to-One: Since f′(x)>0f'(x) > 0, the function is strictly increasing and thus one-to-one.
  • Onto: As a cubic polynomial with a positive leading coefficient, lim⁡x→∞f(x)=∞\lim_{x \to \infty} f(x) = \infty and lim⁡x→−∞f(x)=−∞\lim_{x \to -\infty} f(x) = -\infty. Therefore, the range of the function is RR, which matches the codomain RR. The function is onto.
  • Conclusion for Example 2: One-to-one and onto.