f(x)=(2x-3)√x. Найти f'(1)+f(1)

f(1)=(2-3)√1=-1f`(x)=2√x+(2x-3)/2√x=(4x+2x-3)/2√x=(6x-3)/2√xf`(1)=(6-3)/2√1=3/2=1,5f(1)+f`(1)=-1+1,5=0,5