posted by Pd .
find any stationary points of the function g(x) = (2x-3)square root of 5+16x-4x^2 (0 less than or equal to x less than or equal to 4) and use the first derivative test to classify each stationary point as a local maximum or local minimum of g(x)
this is how far i have got with an answer
g'(x) = (2x-3) (sqrt(5+16x-4x^2) + 2sqrt(5+16x-4x^2)
= (-4x+16)(2x-2)/sqrt(5+16x-4x^2) + 2sqrt(5+16x-4x^2)
-2(8x^2-30x+7) / sqrt(-4x^2+16x+5)
Set this to 0:
0 = -2(8x^2-30x+7) / sqrt(-4x^2+16x+5)
0 = -2(8x^2-30x+7)
0 = -2(2x-7)(4x-1)
Set each factor to 0:
2x-7 = 0, so x = 7/2
4x-1 = 0, so x = 1/4
can't work out how to get the xLEFT and xRIGHT to classify each using the first derivative to get local min or local max...
Use the second derivative. It it is positive at the stationary point the function is about to get bigger so that is a minimum. If it is negative, the function is about to get smaller so that is a maximum.
By the way I did not check what you did. I suspect it is right because you came out with reasonable points where the function is horizontal.
call your tutor rather than trying to get someone else to do your course work for you