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.
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)
Simplify:
-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)
Factor:
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...
help!
2 answers
call your tutor rather than trying to get someone else to do your course work for you
1 answer