What is the question? Are you just supposed to calculate h(6)?
Is your h-1/2 supposed to be h^(-1/2) = 1/sqrt(h) ?
Is the x-1 actually 1/x ?
I looks like all you have to do is plug in 6 for x and do the calculation. That is not calculus.
x = 6
h(x) = (x-1/2 + 5x)(1 - x-1)
2 answers
Im sorry ... it is a derivative question... it says to use the power rule