证明k≠0时f(x)=x^(5/3)-kx^(4/3)+k²x单调递增的方法问询

Got it, let's break this down step by step to prove the derivative is always positive (or non-negative at worst) — exactly what we need to confirm the function's monotonicity.

First, let's recap the setup we're working with:

• We start with the original function: (f(x) = x^{5/3} - kx^{4/3} + k^2x)
• Taking its derivative gives us the expression we need to analyze:
  [
  f'(x) = \frac{5}{3}x^{2/3} - \frac{4k}{3}x^{1/3} + k^2
  ]

Step 1: Simplify with a Substitution #

Fractional exponents can be tricky, so let's use a substitution to turn this into a quadratic (which we know how to analyze easily). Let (t = x^{1/3}) — this means (x^{2/3} = t^2). Plugging this into the derivative gives:
[
f'(x) = \frac{5}{3}t^2 - \frac{4k}{3}t + k^2
]
Multiply the entire expression by 3 (since 3 is positive, this doesn't change the sign of the inequality we're trying to prove):
[
3f'(x) = 5t^2 - 4kt + 3k^2
]

Step 2: Analyze the Quadratic in (t) #

Now we have a quadratic function (g(t) = 5t^2 - 4kt + 3k^2). To show this is always positive, we can use the discriminant of the quadratic:
For a quadratic (at^2 + bt + c), the discriminant is (D = b^2 - 4ac). Here:

• (a = 5), (b = -4k), (c = 3k^2)
  Calculating the discriminant:
  [
  D = (-4k)^2 - 4(5)(3k^2) = 16k^2 - 60k^2 = -44k^2
  ]

Let's break down what this means:

• For any real number (k), (k^2 \geq 0), so (D = -44k^2 \leq 0)
  • If (k = 0), then (g(t) = 5t^2), which is non-negative (equal to 0 only when (t=0), i.e., (x=0)). For all other (x), (g(t) > 0), so (f'(x) > 0).
  • If (k \neq 0), then (D = -44k^2 < 0). Since the coefficient of (t^2) (5) is positive, the quadratic opens upwards and never crosses the t-axis — meaning (g(t) > 0) for all real (t), so (f'(x) > 0) for all (x).

Step 3: Translate Back to the Original Derivative #

Since (3f'(x) = g(t)) is always positive (or non-negative only at the isolated point (x=0) when (k=0)), this confirms (f'(x) > 0) for all (x \neq 0), and (f'(0) = 0) only when (k=0). Either way, the original function (f(x)) is strictly increasing over its domain.

内容的提问来源于stack exchange，提问作者Scavenger23