Find the two x-intercepts of f(x)=x(x−3), and show that f'(x)=0 is at some point between the x-intercept.

Question

1. Given f(x)=x(x2−x−2), find all the values of c in the open interval [−1,1] such that f'(c)=f(1)−f(−1//1−(−1)
2. Find the two x-intercepts of f(x)=x(x−3), and show that f'(x)=0 is at some point between the x-intercept.

Answer 1

1. two x intercepts o f(x)=x(x-3) are 0 and 3, or (0,0) and (3,0)

f(x) = x(x-3) = x^2-3x

f'(x) = 2x-3 = 0

2x= 3

x =3/2

when 2x=3, x = 3/2 = 1.5 which is between x=0 and x=3, exactly half way between the two intercepts

f(x) is an upward opening parabola, with axis of symmetry x=1.5, and vertex (1.5, -2.25)= minimum point on the parabola. It's the point where the tangent line has zero slope, where the derivative is zero.