QUESTION  




Show that f is differentiable at 1 and find f'(1)



SOLUTION: 


we first check the differentiability of f(x)=2x-3 at (0,1), to do this we use the test of differentiability which is 


at (0,1)

and 


Hence at (0,1) the function was differentiable thereby yielding (2,2).
Now to determine the differentiability of 
ƒ(x)=x², by using the test of differentiability, the function becomes indifferentiable at the given interval (1,2], 
Lastly we need to show that f is differentiable at 1, this is pretty simple because we have proved this above already,ƒ(x)=2x-3 is differentiable at 1 but ƒ(x)=x² is not differentiable at 1, we also need to find  f'(1), if f(x)=2x-3 then f'(x)=2 and hence f'(1)  does not exist because f'(x) is a constant, but ƒ(x)=x² and ƒ'(x)=2x, ƒ'(1)=2(1)=2 exists.