f(x)= -2x^(1/2)+x^(3/2), x=4.

So I know we need a point, and a slope.

f(4)= -2(sqrt4) + (sqrt16) with an index of 3.

f(4)= -4+ 2(sqrt2) with an index of 3.

so are point is (4, -4+ 2(sqrt2) with an index of 3)

now we need to find f'(4) to find the slope which is : limit of x approaching 4 = f(x)- f(4)/ (x-4)

I feel like I am not doing something right. I do not think I can find a conjugate with this problem.