3 x^2 -16 x + 5 = 0
first derivative
at x = {16 +/-sqrt(256-60)]/6
or [16+/-14]/6 = 5 or 0.3333
second derivative
6 x - 16
is that + or - at the two points where the slope is zero?
at x = .333
2 - 16 is negative so maximum there
at x = 5
6x - 16 = 30-16 is positive so minimum there
What is the point of inflection of the function f(x)=x^3-8x^2+5x+50? Is the local extrema at f(1/3)=50.8 max and f(5)=0 min?
For the second derivative I got f(x)=6x-16 and then got x=16/6 when set to zero.
1 answer
1 answer