Thank you CPhill
very clear
thank you Alan and heureka
So here can I wrote $$\frac{3}{8}x$$ insted of $$\frac{3}{8}(x-3)$$ ?
Also in this question, why they didn't wrote $$\frac{1}{2}(x+1)$$ insted of $$\frac{1}{2}x$$ !
But why we cant multiply the negative sign inside the parentheses in this step
-(x^2 - 6x + 9) + 9 - 5
so its become
(-x^2 + 6x - 9) + 9 - 5
But where is my mistake here
thank you Alan
and sorry for the picture
thank you alan
This is the model answer
I did this step because when x tends to 0 , $$\frac{1}{e^}\frac{1}{x^2}$$ is tends to zero , so I replaced it by x