Here's my calculation:
(f(2-h) - f(2))/h = (3-(4-4h+h^2)-0)/h = (4h-h^2 -1)/h = 4 - h - 1/h. Let h-->0 and the limit doesn't exist.
To answer the last question, f(c) is the value of f(x) at exactly c, not the limiting value.
Let f(x) = 3-x^2 when -1<=x<2
And 2x-4 when 2<=x<=4
Now in book when they are calculating left hand derivative they do like that
[f (2-h)-f(2)]/h h----0
Here they have not taken f (2) =0(reason being 2 is not in domain of left hand function) so they take f (2) as limit value =-1
Hence they calculate LHD =-4
sir my doubt is that when we define tangent ,we take two point P and Q. in which one point is always fixed (P) and other always try to approach(Q) .so we should take exact value of f(2)=0 ,as f is defined at 2.
ONE MORE IMPORTANT POINT
derivative of f(x) at point c is defined as [ f(x)-f(c)]/(x-c) where x------c.
here f(c) is value of f at exact value of c or limiting value?
Do you have a question about ragnar's post? He is correct. The limit does not exist.
-Dan
Edit: We can calculate the limit of the derivative as it approaches x = 2 from the left hand side. That value is, indeed -4. That may be what your question is after. In that case we would take instead of f(2) in the derivative equation.
why we take limit from negative side at 2 of f(x) instead of f(2).what is the real definition of derivatives .if derivative of f(x) at point c is defined as [ f(x)-f(c)]/(x-c) where x------c.then why definition does not mention these things that f(c) is not a exact value in derivatives
I'm not sure why you would take the limit from the left, but it is sometimes what you are asked to do in a math class. So it all depends on either what the teacher is trying to say in class or what you are asked to do in your homework. If the question is to find the limit as x-->2-, then that's what you do, though you'll still find that the limit doesn't exist for the difference quotient. The limit only exists for the function itself as x-->2-.
I assure you that f(c) is the exact value of f(x) at c.
They had better NOT be doing that. Where ever 2+ h happens to be, the value of f(2), by that definition, is 0.
The derivative (at x= 2), from the left is
\lim_{h\to 0^-}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0}\frac{3- (2+ h)^2 - 0}{h}= \lim_{h\to 0}\frac{3- 4- 4h+ h^2}{h}= \lim_{h\to 0}\frac{-1- 4h+ h^2}{h}
Now, the denominator goes to 0 (as it does in all such derivative limits) but the numerator does not- the left side limit does not exist so the limit itself, and therefore the derivative, does not exist at x= 2.
In fact, we really did not need to do that calculation.
\lim_{x\to 2^-} f(x)= \lim_{x\to 2} 3- (\lim_{x\to 2} x)^ 2- 4= -1\ne f(2)= 0
so the function is not even continuous at x= 2 and can't be differentiable there.
Absolutely. You are completely correct. Whoever told you to take \lim_{x\to 2^-}f(x)= -1 was wrong!sir my doubt is that when we define tangent ,we take two point P and Q. in which one point is always fixed (P) and other always try to approach(Q) .so we should take exact value of f(2)=0 ,as f is defined at 2.
f(c) is the value of f at x= c. It is NOT a limiting value.ONE MORE IMPORTANT POINT
derivative of f(x) at point c is defined as [ f(x)-f(c)]/(x-c) where x------c.
here f(c) is value of f at exact value of c or limiting value?