Given the function

Question

y= 1/3x³+x²-35x+10

i)Determine the critical values and find out whether these critical values are maxima or minima

 

ii) Determine the extreme values (y values) at the maxima and minima of the function

Answer 1

If y= 1/3x³+x²-35x+10 then y' = x² + 2x -35 = (x+7)(x-5) and setting y' = 0 gives critical x values 5 and -7.

 

Since y" = 2x + 2, we have y"(-7) < 0 so concavity is downward and a relative maximum exists when x=-7.

Also, y"(5) > 0 so concavity there is upward and thus there's a relative minimum when x=5.