Question

Use both the first and second derivative tests to show that f(x) = 3x2 − 6x + 1 has a relative minimum at x=1.

Answer 1

Explanation:

Given: f(x) = 3x^2 − 6x + 1. x ∈ (-∞, ∞)

now differentiating f(x) w.r.t. x we get

f'(x) = 6x-6

equating it 0 we get

x = 1, which is a critical point

Now using second derivative test to check relative maxima or minima

f''(x) = 6 >0

⇒ x=1 is a relative minima.

also, x=1 f(1) = 3-6+1 = -2