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Derivatives: Handling Multi-Variable Always-True Inequalities

High school math notes on derivatives, using an example inequality involving e^x to demonstrate three methods: parameter separation, Taylor expansion, and derivatives, for handling multi-variable always-true problems.

Machine-translated from the Chinese original.

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Approach

  1. Separation (full separation / partial separation)
  2. Case analysis (use special points to narrow the range)
  3. Special value →\rightarrow necessity →\rightarrow sufficiency
  4. Use bounding →\rightarrow sufficiency →\rightarrow necessity (proof by contradiction)

A Good Problem

Let f(x)=ex−1−x−ax2f(x)=e^x-1-x-ax^2.
(I) If a=0a=0, find the monotonic intervals of f(x)f(x).

Monotonically decreasing on (−∞,0)(-\infty,0), monotonically increasing on (0,+∞)(0,+\infty); steps omitted.

(II) If f(x)≥0f(x)\ge 0 for x≥0x\ge 0, find the range of values for aa.

Method 1

f(x)≥0f(x)\ge 0 is equivalent to ex−1−x≥ax2e^x-1-x\ge ax^2

(1) x=0x=0: holds

(2) x>0x\gt 0: a≤ex−x−1x2a\le\frac{e^x-x-1}{x^2}

Let g(x)=ex−x−1x2g(x)=\frac{e^x-x-1}{x^2}, then g′(x)=(x−2)ex+x+2x3g'(x)=\frac{(x-2)e^x+x+2}{x^3}

Let t(x)=(x−2)ex+x+2t(x)=(x-2)e^x+x+2, then t′(x)=(x−1)ex+1t'(x)=(x-1)e^x+1, t′′(x)=xext''(x)=xe^x

∵t′′(x)=xex>0\because t''(x)=xe^x\gt 0
∴t′(x)\therefore t'(x) is monotonically increasing, t′(x)>t′(0)=0t'(x)\gt t'(0)=0
∴t(x)\therefore t(x) is monotonically increasing, t(x)>t(0)=0t(x)\gt t(0)=0, g′(x)>0g'(x)\gt 0
∴g(x)>lim⁡x→0g(x)\therefore g(x)\gt\lim_{x\to 0}g(x)

lim⁡x→0g(x)=lim⁡x→0ex−x−1x2\lim_{x\to 0}g(x)=\lim_{x\to 0}\frac{e^x-x-1}{x^2}

By L’Hopital’s rule backup

=lim⁡x→0ex−12x=\lim_{x\to 0}\frac{e^x-1}{2x} =lim⁡x→0ex2=\lim_{x\to 0}\frac{e^x}2 =12=\frac 1 2 ∴a≤12\therefore a\le\frac 1 2

Method 2

Taylor expansion (background)

Idea

∵ex=1+x+x22!+x33!+...\because e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+... ∴ex≥1+x+ax2⇒a≤12\therefore e^x\ge 1+x+ax^2\Rightarrow a\le \frac 1 2

Execution

We can prove ex≥1+x+12x2e^x\ge 1+x+\frac 1 2 x^2; sufficiency is obvious, but necessity is hard to establish this way.

Method 3

Derivatives

Idea

f(x)≥f(0)=0f(x)\ge f(0)=0; intuitively, f′(x)f'(x) should be greater than or equal to 00 near x=0x=0.

f′(x)=ex−2ax−1f'(x)=e^x-2ax-1, f′(0)=0f'(0)=0, so f′′(x)f''(x) should be greater than or equal to 00 near x=0x=0.

f′′(x)=ex−2af''(x)=e^x-2a, so f′′(0)≥0f''(0)\ge 0, i.e. a≤12a\le\frac 1 2 (necessary)

Execution

Suppose a>12a\gt\frac 1 2, then f′′(0)<0f''(0)\lt 0, so there exists an interval near x=0x=0 where f′′(x)<0f''(x)\lt 0; call it [0,x0)[0,x_0).

∵f′′(x)<0,x∈[0,x0)\because f''(x)\lt 0,x\in[0,x_0)
∴f′(x)\therefore f'(x) is monotonically decreasing on x∈(0,x0)x\in(0,x_0), so f′(x0)<f′(0)=0f'(x_0)\lt f'(0)=0, and f′(x)<0,x∈(0,x0)f'(x)\lt 0,x\in(0,x_0)
∵f′(x)<0,x∈(0,x0)\because f'(x)\lt 0,x\in(0,x_0)
∴f(x)\therefore f(x) is monotonically decreasing on x∈[0,x0)x\in[0,x_0), so f(x0)<f(0)=0f(x_0)\lt f(0)=0, which contradicts the problem’s requirement
∴a≤12\therefore a\le\frac 1 2 (necessary)

Combining Method 2 and Method 3 proves the necessary-and-sufficient condition.

Method 4

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