# function problem

• Nov 11th 2012, 10:23 AM
sunrise
function 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
• Nov 11th 2012, 10:29 AM
richard1234
Re: function problem
C (although it is neither increasing nor decreasing at x = 0)

All the other ones are decreasing for part of the domain.
• Nov 11th 2012, 10:31 AM
Plato
Re: function problem
Quote:

Originally Posted by sunrise
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?

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.
• Nov 11th 2012, 10:58 AM
sunrise
Re: function problem
richard,
How you solved and thought that C is increasing?
Should i give value to the x?
• Nov 11th 2012, 11:52 AM
richard1234
Re: function problem
Quote:

Originally Posted by sunrise
richard,
How you solved and thought that C is increasing?
Should i give value to the x?

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).
• Nov 11th 2012, 12:13 PM
Plato
Re: function problem
Quote:

Originally Posted by richard1234
(C) is always increasing (except when x = 0, where the derivative is zero).

Technically the function \$\displaystyle f(x)=3x^3-11\$ is increasing everywhere.

The the statement that \$\displaystyle f\$ is an increasing function means that if \$\displaystyle a<b\$ then \$\displaystyle f(a)<f(b)\$.

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 \$\displaystyle a<b\$ then \$\displaystyle a^3<b^3\$ then \$\displaystyle a^3-11<b^3-11\$. Proved.
• Nov 11th 2012, 12:14 PM
sunrise
Re: function problem
If we think from your idea,

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

Thanks
• Nov 11th 2012, 12:17 PM
Plato
Re: function problem
Quote:

Originally Posted by sunrise
If we think from your idea,
\$\displaystyle f(x) = 2x^2 + 9\$ is also increasing isnt it?

No it is not.
\$\displaystyle -3<1\$ BUT \$\displaystyle f(-3)>f(1)\$.
• Nov 11th 2012, 12:22 PM
sunrise
Re: function problem
Now I understand thank you.