How is 1/f(x) differentiable at 0.

In summary: Therefore, it is differentiable at 0. In summary, the conversation discusses the differentiability of 1/f(x) at x=0. The speaker proves that the derivative exists at 0, despite f(x) not being equal to 0, using the definition of differentiability and the given function f(x). The other person's confusion is clarified and it is stated that a function can be differentiable at a point even if it is not equal to 0. An example is given to illustrate this concept.
  • #1
asif zaidi
56
0
I can prove derivative of 1/f(x) = -f'(x)/(f^2).
But how can this be differentiable at x=0 (if f(x) is not equal to 0) as my class notes claim

Thanks
 
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  • #2
Say f(0) = 2. This implies 1/f(0) = 1/2 and [1/f'(0)]' = -f'(0)/4.
 
Last edited:
  • #3
True but let me tell you my confusion ( should have posted in detail)

I proved 1/f'(x) as follows

- given a seq {an} -> a,

1/f(an) - 1/f(a) / (an -a) -> a then this function is differentiable.

Manipulating this:

( f(a) - f(an) ) /( f(a)f(an) (an-a) =

-f'(a) / ( f(a) f(an) ).

Since an-> a, f(an)-> f(a) and thus I get -f'(a)/f^2.

But this is my problem- I say an-> a. But 'a' can be any value including 0 but yet it is give that f(x) is not equal to 0.

Hope I am making sense

Thanks

Asif
 
  • #4
f(x) is not equal to zero, that in no way implies that a may not be zero.
 
  • #5
A function is differentiable at a point if the derivative exists at a point. From the derivative you got, the derivative exists at 0, as seen from just plugging in 0 and knowing that [tex]f(x) \neq 0[/tex]. Thus it is differentiable at 0. That is all.
 
  • #6
Example: f(x) = 2 + x. In particular, f(0) = 2 > 0. And, f'(x) = 1 for all x.

Then, derivative of 1/f(0) = -f'(0)/f(0)^2 = -1/4.
 

What does it mean for 1/f(x) to be differentiable at 0?

Differentiability at a point means that the function is smooth and has a well-defined tangent line at that point. In other words, the function can be approximated by a straight line at that point.

Why is differentiability at 0 important for 1/f(x)?

Differentiability allows us to make predictions about the behavior of the function at that point and in its neighborhood. It also allows us to use calculus techniques such as finding the derivative and solving optimization problems.

How can we determine if 1/f(x) is differentiable at 0?

We can use the definition of differentiability, which states that a function is differentiable at a point if the limit of the difference quotient exists at that point. In other words, if the left and right-handed limits of the difference quotient are equal, then the function is differentiable at that point.

What happens if 1/f(x) is not differentiable at 0?

If the function is not differentiable at 0, it means that the function is not smooth at that point and it does not have a well-defined tangent line. This could be due to a sharp corner or a discontinuity at that point.

Can 1/f(x) be differentiable at 0 but not continuous at 0?

No, if a function is differentiable at a point, it must also be continuous at that point. However, the opposite is not necessarily true. A function can be continuous at a point but not differentiable at that point.

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