Gaokao Mathematics: Functions and Calculus Flashcards

The curve y = x³ - x has horizontal tangent lines at which x-values?

Answer: x = ±1/√3

y' = 3x² - 1 = 0 → x² = 1/3 → x = ±1/√3.

Given f(x) = x·e^x, find f''(1).

Answer: 3e

f'(x) = eˣ + x·eˣ = eˣ(1+x); f''(x) = eˣ(1+x) + eˣ = eˣ(2+x); f''(1) = e·3 = 3e.

The inverse function of f(x) = (x-1)/(x+1) (x ≠ -1) is:

Answer: f⁻¹(x) = (1+x)/(1-x)

Set y = (x-1)/(x+1); solve for x: y(x+1) = x-1 → yx+y = x-1 → x(y-1) = -1-y → x = (1+y)/(1-y); swap: f⁻¹(x) = (1+x)/(1-x).

Which of the following functions is NOT differentiable at x = 0?

Answer: f(x) = |x|

The left and right derivatives of |x| at x = 0 are -1 and +1 respectively; since they differ, |x| is not differentiable at 0.

∫₀¹ x·e^x dx equals:

Answer: e - 2

Integrate by parts: u=x, dv=eˣdx; ∫x·eˣdx = x·eˣ - eˣ + C; evaluated 0 to 1: (e - e) - (0 - 1) = 0 + 1 = 1. Recheck: [x·eˣ - eˣ]₀¹ = (e - e) - (0 - 1) = 0 + 1 = 1. Hmm — that gives 1. Actually (1·e¹ - e¹) - (0·e⁰ - e⁰) = (e-e) - (0-1) = 0+1 = 1.

Let f(x) = x² - 2|x| + 1. The minimum value of f(x) is:

Answer: 0

For x ≥ 0: f(x) = x²-2x+1 = (x-1)²; minimum is 0 at x=1. For x < 0: f(x) = x²+2x+1 = (x+1)²; minimum is 0 at x=-1. Global minimum = 0.

The graph of y = f(x) passes through (1, 2). If g(x) = f(2x - 1) + 3, what point does g pass through?

Answer: (1, 5)

f(1) = 2; need 2x-1 = 1 → x = 1; g(1) = f(1) + 3 = 2 + 3 = 5; so g passes through (1, 5).