f(x) = (x + 1)/(x − 1) is the function. I've tried and I keep getting dy/dx as -2/(x-1)^2 through the quotient rule.
But that's not equal to f(x) which is what I was asked to prove. This is an examination question if that helps.
Can someone show me step-by-step how to do this properly?
How to prove this function's f'(x) and f(x) are the same?
2016-06-20 7:42 am
回答 (2)
2016-06-20 7:52 am
If f(x) = g(x)/h(x)
then f'(x) = [h(x)g'(x) - g(x)h'(x)] / [h(x)]²
f(x) = (x + 1)/(x - 1)
Then, f'(x)
= [(x - 1)(x + 1)' - (x + 1)(x - 1)'] / (x - 1)²
= [(x - 1) - (x + 1)] / (x - 1)²
= -2 / (x - 1)²
Then f'(x) ≠ f(x)
(Your calculations are correct.)
then f'(x) = [h(x)g'(x) - g(x)h'(x)] / [h(x)]²
f(x) = (x + 1)/(x - 1)
Then, f'(x)
= [(x - 1)(x + 1)' - (x + 1)(x - 1)'] / (x - 1)²
= [(x - 1) - (x + 1)] / (x - 1)²
= -2 / (x - 1)²
Then f'(x) ≠ f(x)
(Your calculations are correct.)
2016-06-20 7:52 am
You are correct that dy/dx=-2/(x-1)^2 and the only way for
-2/(x-1)^2 = (x+1)/(x-1) is if -2=(x-1)(x+1) giving x^2=-1 which gives impossible
values of x. If it is an exam question then there is an error. The only function
such that f'(x)=f(x) is f(x)=Ae^x where A is a constant .
-2/(x-1)^2 = (x+1)/(x-1) is if -2=(x-1)(x+1) giving x^2=-1 which gives impossible
values of x. If it is an exam question then there is an error. The only function
such that f'(x)=f(x) is f(x)=Ae^x where A is a constant .
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