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# C3 differentiation help? Watch

1. The function f is defined for x > 0 by

f(x) = e^(-2x) + 3/x +3

(ii) Hence prove that f is a decreasing function

----

I have the differential as f'(x) = -2e^(-2x) -3/x^2, how do I show it's decreasing? I know that if you substitute x = 0 in then you will get an error. Do I substitute in x=1, and show that it's decreasing? Or should I say as x->infinity then f'(x) -> -infinity?

2. (Original post by TobyReichelt)
The function f is defined for x > 0 by

f(x) = e^(-2x) + 3/x +3

(ii) Hence prove that f is a decreasing function

----

I have the differential as f'(x) = -2e^(-2x) -3/x^2, how do I show it's decreasing? I know that if you substitute x = 0 in then you will get an error. Do I substitute in x=1, and show that it's decreasing? Or should I say as x->infinity then f'(x) -> -infinity?

Pull out a factor of -1. That gets you

Since the inside of the brackets is always positive (because you're adding together two positive terms) then the negative outside means f'(x) is always negative. So decreasing.
3. (Original post by TobyReichelt)
The function f is defined for x > 0 by

f(x) = e^(-2x) + 3/x +3

(ii) Hence prove that f is a decreasing function

----

I have the differential as f'(x) = -2e^(-2x) -3/x^2, how do I show it's decreasing? I know that if you substitute x = 0 in then you will get an error. Do I substitute in x=1, and show that it's decreasing? Or should I say as x->infinity then f'(x) -> -infinity?

Think about each term of the derivative separately. What can you say about x^2? what about e^-2x? (try writing this in a slightly different way) also notice how you can factor out the -1...
4. (Original post by Zacken)
Pull out a factor of -1. That gets you

Since the inside of the brackets is always positive (because you're adding together two positive terms) then the negative outside means f'(x) is always negative. So decreasing.
Thank. You. So. Much.
5. (Original post by TobyReichelt)
Thank. You. So. Much.
No worries.

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Updated: September 10, 2016
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