Prove or disprove this statement.

Question

f(t) = t³-15t+c has at most one root in the interval [-2,2] where c is constant

Answer 1

f(t) = t³ - 15t + c

f'(t) = 3t² - 15 = 3(t² - 5)

f'(t) = 0 when t = ±√5 ≈ ±2.23

When t < -√5, f'(t) > 0. So, f(t) is strictly increasing on (-∞, -√5)

When -√5 < t < √5, f'(t) < 0. So, f(t) is strictly decreasing on (-√5, √5)

When t > √5, f'(t) > 0. So, f(t) is strictly increasing on (√5, ∞)

Since f(t) is strictly decreasing between -√5 and √5, f(t) is also strictly decreasing between -2 and 2.

So, the graph of f(t) can cross the x-axis at most once between x = -2 and x = 2.