Question

How do I prove a function has a horizontal tangent line?

Answer 1

1) How can you find a tangent line?
2) How do you know if a line is horizontal?

Answer 2

take the derivative and set it equal to zero, plug in for x

Answer 3

to find a point on the tangent line, which would prove it exists...

Answer 4

Finding a point on the curve where derivative is zero would suffice, I suppose.

Answer 5

Everything sounded ok except for that last part you said:

"take the derivative and set it equal to zero, `plug in` for x"

what you're looking to do is,

"take the derivative and set it equal to zero, `solve` for x"

I think you have the right idea, you just said it a little strangely.

Answer 6

\[\frac{ 1 }{ x }(x^2+e^x)\] What if it doesn't come out nice?

Answer 7

Is that the function you need to prove has a horizontal tangent?

Answer 8

yes

Answer 9

hmm

Answer 10

If the derivative does not turn out 'nicely', it might come down to 'logicking' it out.
Like, by observation we should get some deal of a fraction since we are starting with a fraction. That means we will have some R(x) / Q(x) = 0.

We should recall, then, that any fraction is equal to 0 if and only if its numerator is equal to zero (and its denominator is nonzero), so we would be able to check: R(x) = 0 (and whatever answer from that in Q(x) for bad types of zeros)

Answer 11

Have we figured out the derivative of the function? We can play with it from there. :)

Answer 12

I think it is ... \[\frac{ x^2+e^x(x-1) }{ x^2 }\]

Answer 13

Yep, that appears correct to me. :)
So, we would set this equal to zero.
\( \displaystyle \frac{x^2 + e^x (x - 1)}{x^2} = 0 \)

Answer 14

Yup, I got that far lol

Answer 15

If we then restrict \(x \neq 0\), we could simply take the numerator.
\(x^2 + e^x (x - 1) = 0\)

Answer 16

So, I cannot think of many ways to manipulate this into any more useful form... so I would consider either a calculator, or there might be some process to estimate it. Hm...

Answer 17

My first question to myself is, "how does the function behave at simple points? 0? 1? going to infinity?"

For x=0;
x^2 + e^(x) (x - 1)
0^2 + e^(0) (0 - 1)
1 * -1 = -1 Is a negative value.
For x=1
1^2 + e^(1) (1 - 1) = 1 + 0 = 1 Is a positive value

So there is definitely a root somewhere between 0 and 1.

As x approaches infinity,
x^2 + e^x (x - 1) diverges to infinity, quite simply. x^2 *woosh squared* e^(x) *exponential woosh*, times more infinity *woosh*

Answer 18

Ok, that makes sense...

Answer 19

For x = -1 and -infinity:
x=-1
(-1)^2 + e^(-1) (-1 - 1)
1 + e^(-1) * -2
1 - 2e^(-1) 2e^(-1) is roughly 2/2.7, so its less than 1
so its some positive value, which means there is also a root between 0 and -1.

x--> -infinity
x^2 goes to infinity, while e^(x) goes to 0 and (x - 1) goes to -infinity. However, the exponential should increase much faster than the linear x and pretty much cancels the effect of it. So this diverges to infinity...

Answer 20

This is roughly what it would look like, then, and we just need to find the two in-between points.

I'd say at this point, if you can use a calculator, this is that time. :P
If we have to do it by hand, we need an estimating technique, which I cannot think of...