Question

f(x)=-3 va rog ; f(x)=1 pe 2 x-3

Answer 1

2x-3=-3
2x=0
x=0


2x-3=1
2x=1+3=4
x=2

Answer 2

$F(x)= \int\limits { \frac{1}{2x-3} } \, dx = \frac{1}{2} \int\limits \frac{(2x-3)'}{2x-3} x} \, dx= \frac{1}{2}ln/2x-3/+C \\ F(x)=-3 \\ \frac{1}{2}ln/2x-3/+C=-3 \\ C=-3-ln \sqrt{2x-3}$