Gaokao Mathematics: Functions and Calculus Flashcards

The function f(x) = x/(1+|x|) has range:

Answer: (-1, 1)

For x ≥ 0: f = x/(1+x) ∈ [0,1); for x < 0: f = x/(1-x) ∈ (-1,0); overall range (-1,1), open endpoints.

Given f(x) = x⁴ - 4x, find the value of x at which f attains its minimum on ℝ.

Answer: x = 1

f'(x) = 4x³ - 4 = 4(x³-1) = 0 → x = 1; f''(1) = 12 > 0, confirming minimum.

Using the limit definition, which expression correctly represents f'(a)?

Answer: lim_{h→0} [f(a+h) - f(a)] / h

The derivative is defined as the limit of the difference quotient as h → 0.

If f(x) = x²·sin(1/x) for x ≠ 0 and f(0) = 0, is f differentiable at x = 0?

Answer: Yes, f'(0) = 0

f'(0) = lim_{h→0} h²sin(1/h)/h = lim h·sin(1/h) = 0 (squeeze: |h·sin(1/h)| ≤ |h| → 0).

The equation of the normal line to y = x² at (1, 1) is:

Answer: y = -x/2 + 3/2

Tangent slope at (1,1): y'=2x → 2; normal slope = -1/2; y - 1 = -1/2(x-1) → y = -x/2 + 3/2.

Evaluate lim_{x→0} (e^x - 1 - x) / x².

Answer: 1/2

Applying L'Hôpital twice: first (eˣ-1)/2x → limit 1/2; or use Taylor: eˣ = 1+x+x²/2+… so (eˣ-1-x)/x² → (x²/2)/x² = 1/2.

The function f(x) = x³ - 6x² + 9x + 2 has local extrema at:

Answer: x = 1 (max) and x = 3 (min)

f'(x)=3x²-12x+9=3(x-1)(x-3); f'=0 at x=1,3; f''(1)=6-12=-60 (local min).