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

Step-by-step 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

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

kk.bj.

step-by-step explanation:

.mnm.kn.kj