Suppose f(t)=6t^3+7t^2+3t−4. Find the value of t in the interval [4,10] where f(t) takes on its minimum value.

Question

 Suppose f(t)=6t^3+7t^2+3t-4. Find the value of t in the interval [4,10] where f(t) takes on its minimum value.

Answer 1

Extrema occur where derivative is zero.

0 =f'(t) = 18t^2 + 14t + 3

t = [ (-14) +or- sqrt( 14^2 - 4(18)(3) ]/36

= [ -14 +or- sqrt( 196 - 216)]/36

= [ -14 +or- sqrt(-20)]/36

= [ -14 +or- 2isqrt(5)]/36

= [-7 +or- isqrt(5)]/18

It's not going to happen