Find the x-coordinate(s) of any inflection point(s) of the graph of the function.?

Question

F(x)= -4x^2+3xlnx

Answer 1

At inflection point: F' ' (x) = 0

f'(x) = -8x + 3 lnx + 3

f' ' (x) = -8 + 3/x

=> f'' (x) = 0 <=> -8 + 3/x = 0 => 3/x = 8
=> x = 3/8 (x>0 so as for lnx to be valid)

f(3/8) = - 4*9/64 + 3 * 3/8 *ln(3/8)
= - 9/16 + 9/8 ln(3/8)
= 9/8( -1/2 + ln(3/8))

So the only inflection point is: (x = 3/8, y = 9/8( -1/2 + ln(3/8)))

Answer 2

F´(x) =-8x+3(lnx +1)
F´´(x) =-8 +3/x
F´´(x) =0 3x=8 and x=3/8
At x= 3/8 F´´ changes sign so there is an inflection point