Hi, I have a function question may you help me?

Which one is increasing function for all x values?

(a) y = |x - 7|

(b) y = 2x^2 + 9

(c) 3x^3 -11

(d) 4x^4 + 2 how can i prove?

Thank you

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- November 11th 2012, 11:23 AMsunrisefunction problem
Hi, I have a function question may you help me?

Which one is increasing function for all x values?

(a) y = |x - 7|

(b) y = 2x^2 + 9

(c) 3x^3 -11

(d) 4x^4 + 2 how can i prove?

Thank you - November 11th 2012, 11:29 AMrichard1234Re: function problem
C (although it is neither increasing nor decreasing at x = 0)

All the other ones are decreasing for part of the domain. - November 11th 2012, 11:31 AMPlatoRe: function problem
You posted this in the pre-calculus forum.

Thus there is no way to prove this one way or the other.

You can simply draw the graphs and see the answer. But that is hardly a poof.

On the other hand, with calculus we can see which one has a non-negative derivative. That would prove it. - November 11th 2012, 11:58 AMsunriseRe: function problem
richard,

How you solved and thought that C is increasing?

Should i give value to the x? - November 11th 2012, 12:52 PMrichard1234Re: function problem
Assigning a value to x has no indication of whether a function is increasing or not.

Why don't you just graph the function? To rigorously prove it, we note that its derivative is 9x^2, which is always non-negative. Therefore the function in (C) is always increasing (except when x = 0, where the derivative is zero). - November 11th 2012, 01:13 PMPlatoRe: function problem
Technically the function

**is increasing everywhere**.

The the statement that is*an increasing function*means that if then .

That is clearly true in this case. When I made the remark reply #3 about proof, it was addressing the fact that this is a precalculus forum.

There is of course a perfectly good way of proving this.

Suppose that then then . Proved. - November 11th 2012, 01:14 PMsunriseRe: function problem
If we think from your idea,

y = 2x^2 + 9 is also increasing isnt it?

Thanks - November 11th 2012, 01:17 PMPlatoRe: function problem
- November 11th 2012, 01:22 PMsunriseRe: function problem
Now I understand thank you.