A particle's position along the x-axis is given by x(t) = t^4/24 - t^3/2 + 2t^2 - 1. What is the maximum acceleration on the interval 0 <= t <= 4?

Question

I found the acceleration to be (t^2 - 6t + 8)/2 and the derivative of the acceleration to be t - 3. I thought the maximum should occur at t = 3, but the second derivative of the acceleration is positive, so that would be a minimum. I'm not sure what to do here.

Answer 1

I agree with you.
I also found the 2nd derivative to be (1/2)t^2 - 3t + 4
and the third derivative to be t - 3

since the 2nd derivative is a quadratic opening upwards, you would have a minimum at t = 3

Could be a misprint