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# differentiable function

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1. 1.
Let
f(x) = 2x +3 x<=2
f(x) = x^2 - 2x + 7 x>2

Show that f'(x) exists at x=2, but f''(2) doesnt exist.

2.
Let f,g be functions s.t. the product h(x)=f(x)g(x) is differentiable at a. Does it follow that f and g are differentiable at a?
is it possible that only 1 of f and g are differentiable at a?

for 1. do u just diff once and show f'(2)=2 so it exists and f''(2)=0 so it doesnt exist?? if so then y is there a need for 2 functions wen we can fork it out from just the first.
2. 1) f'(x) = 2 (x<=2) ->f'(2) = 2
f'(x) = 2x - 2 (x>2) -> f'(2) = 2.
So f'(x) at x= 2 exists
f''(x)= 0 (x<=2)--> f''(2) = 0
f''(x) = 2 (x>2)-->f''(2) = 2 <> 0.
So f''(2) doesn't exist.
2. No, f and g may not be differentiable at a.
Yes (not so sure).
3. Suppose that f'(a) isnt defined, and g(a) = 0 then we have
h'(a) = f'(a)g(a) + f(a)g'(a) = f(a)g'(a) which is defined.
4. k i understand 1 but i dnt get numba 2
6. (Original post by JamesF)
But not enough, I think. What if both g and f are not differentiable? U just proved that 1 of them is differentiable.
7. It asks, does the differentiability of h(x) at a imply that both f(x) and g(x) are differentiable at a, and i showed that the answer is no.
It doesnt take much to extend the arguement to show that neither have to be differentiable at a.
If g(a) = f(a) = 0, then h'(a) is defined (and equals 0) and neither of f or g need be differentiable at a.
8. (Original post by JamesF)
It asks, does the differentiability of h(x) at a imply that both f(x) and g(x) are differentiable at a, and i showed that the answer is no.
It doesnt take much to extend the arguement to show that neither have to be differentiable at a.
If g(a) = f(a) = 0, then h'(a) is defined (and equals 0) and neither of f or g need be differentiable at a.
I might misunderstood the question. I thought it asked if h(x) = f(x).g(x), and was h(x) differentiable if only one of g(x) or f(x) was defined or none of them.
But ur answer for none of them, I think it's wrong. Let's consider this
f(x) = (x-1)^1/2 -> f(1) = 0, f'(1) is not defined.
g(x) = (x-1)^1/3 -> g(1) = 0, g'(1) is not defined.
h(x) = (x-1)^5/6 -> h'(x) = 5/6(x-1)^(-1/6) which is not defined at x = 1.
But quite similarly, ur answer is right if g(x) = (x-1)^2/3. I mean ur answer for neither of them is differentiable can't be generalization.
Updated: January 24, 2005
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