Thread: Graphing Question With Constants A,K

1. Graphing Question With Constants A,K

I have to find where y is increasing/decreasing, where it's concave up/down and any inflection points, if there are horizontal asymptotes, and then sketch it. Here's the problem:

y = x^k/1+x^k

k is a positive integer greater than one. Here's what I get for the first derivative, after that I'm stuck:

f'(x)= kx^(k-1)/(1+x^k)^2

My guess is that the function is increasing for x>o and decreasing for x<0 but stuck after that.

2. Hello,

Originally Posted by SportfreundeKeaneKent
I have to find where y is increasing/decreasing, where it's concave up/down and any inflection points, if there are horizontal asymptotes, and then sketch it. Here's the problem:

y = x^k/1+x^k

k is a positive integer greater than one. Here's what I get for the first derivative, after that I'm stuck:

f'(x)= kx^(k-1)/(1+x^k)^2

My guess is that the function is increasing for x>o and decreasing for x<0 but stuck after that.
This is not true...

If k is odd, then $k-1$ is even $\rightarrow k-1=2k'$. Therefore, $x^{k-1}=x^{2k'}=(x^{k'})^2$, which is always positive !

So you have to differentiate the possibilities "k is odd" & "k is even".

3. I don't really understand. I'm differentiating with respect to x. So I wanna find the derivative for x^k to find where the function increases/decreases and then find the derivative of that to find where it's concave up/down.

I thought the derivative of x^k would be kx^k-1. I dunno what the second derivative would be.

4. Originally Posted by SportfreundeKeaneKent
I don't really understand. I'm differentiating with respect to x. So I wanna find the derivative for x^k to find where the function increases/decreases and then find the derivative of that to find where it's concave up/down.

I thought the derivative of x^k would be kx^k-1. I dunno what the second derivative would be.
But you can't know if $x^{k-1}$ is actually positive or negative. It depends on both k and x.
If x>0, there is no problem.

If x<0, it will be positive if k-1 is positive. Otherwise, it will be negative.

For the second derivative, it's the same thing. You have $k \cdot x^{k-1}$, which has $k(k-1)x^{k-2}$ as derivative...

5. Originally Posted by SportfreundeKeaneKent
I have to find where y is increasing/decreasing, where it's concave up/down and any inflection points, if there are horizontal asymptotes, and then sketch it. Here's the problem:

y = x^k/1+x^k
Next time, please use parentheses/brackets or latex.

I'm guessing you mean this:

$y=\frac{x^k}{1+x^k}$

So, using the quotient rule to find the derivative:

$y'=\frac{(\frac{d}{dx}x^k)(1+x^k)-[\frac{d}{dx}(1+x^k)](x^k)}{(1+x^k)^2}$

Derivate the terms on the right:

$y'=\frac{(kx^{k-1})(1+x^k)-kx^{k-1}(x^k)}{(1+x^k)^2}$

Simplify a little bit:

$y'=\frac{kx^{k-1}+kx^{k-1}(x^k)-kx^{k-1}(x^k)}{(1+x^k)^2}$

$y'=\frac{kx^{k-1}}{(1+x^k)^2}$

Convert the form to make your life easier for the next derivative:

$y'=kx^{k-1}(1+x^k)^{-2}$

Now take the next derivative using the product rule:

$y''=kx^{k-1}[\frac{d}{dx}(1+x^k)^{-2}]+(1+x^k)^{-2}[\frac{d}{dx}kx^{k-1}]$

Derivate the first term using the chain rule:

$y''=kx^{k-1}(-2)(1+x^k)^{-3}kx^{k-1}+(1+x^k)^{-2}[\frac{d}{dx}kx^{k-1}]$

And then the second term:

$y''=kx^{k-1}(-2)(1+x^k)^{-3}kx^{k-1}+(1+x^k)^{-2}k(k-1)x^{k-2}$

Simplify a bit:

$y''=-2k^2x^{2k-2}(1+x^k)^{-3}+(1+x^k)^{-2}k(k-1)x^{k-2}$

That should get you started.