Question

Does anyone use wolframalpha need help

Answer 1

With?

Answer 2

How would i write this in wolfram alpha

Answer 3

The developer's haven't implemented a way to interpret every piecewise function just yet. For the moment, you can use the syntax used in Mathematica (software developed by the parent company), which would be
`Piecewise[ { {-x^2 + 2, x != 2}, {-5, x == 2} } ]`
See the result here:
http://www.wolframalpha.com/input/?i=Piecewise%5B+%7B+%7B-x%5E2+%2B+2,+x+!%3D+2%7D,+%7B-5,+x+%3D%3D+2%7D+%7D+%5D&dataset=

Answer 4

Thank you, is that one exclamation mark suppose to be a one

Answer 5

No, `!=` is used to indicate "not equal" (while a double equal sign `==` means "equal").

Answer 6

Oh thank you

Answer 7

Dang i was hoping it would show the step by step thing, but can you help me with this?

Answer 8

It also works with `=/=` if that's easier to remember.

Answer 9

Steps for what?

Answer 10

For the equation

Answer 11

I dont know how to do it

Answer 12

Do you mean steps for how to plot the function?

Answer 13

No how to solve it, or can you solve it? Or is it a graph thing?

Answer 14

There's nothing to solve here...

Answer 15

Oh hold on, I didn't notice the limit. Is that what you're talking about?

Answer 16

Yes i think so

Answer 17

Lol yeah i just tried to enter the limit into wolfram alpha

Answer 18

You can get the limit with "limit as x approaches 2 of ..." where "..." is the `Piecewise[]` input. I doubt steps will work though.

I always assume they don't so that I'm pleasantly surprised when they do.

Answer 19

Haha, ok thanks so i will try to figure it out, but is this the same as what the equations asks?

Answer 20

http://www.wolframalpha.com/input/?dataset=&i=lim+f(x)+%3D+Piecewise%5B+%7B+%7B-x%5E2+%2B+2,+x+!%3D+2%7D,+%7B-5,+x+%3D%3D+2%7D+%7D+%5D+as+x-%3E2

Answer 21

Yeah, the problem is to find the limit of the function defined by the piecewise \(f(x)\). The "input interpretation" is there to let you know that what you typed into the input field is interpreted properly.

Answer 22

Ok thanks

Answer 23

Anyway, the thing about "the limit of \(f(x)\) as \(x\to2\)" is that you're considering the behavior of \(f(x)\) *near* \(x=2\), but not *at* \(x=2\).

Your function is conveniently defined so that \(f(x)=-x^2+2\) for every value of \(x\) except \(2\). This ultimately means that
\[\lim_{x\to2}f(x)=\lim_{x\to2}(-x^2+2)=\cdots\]

Answer 24

So all i would ahve to do is find out that -2 is closes to the function that represent 2 or closest to the number 2

Answer 25

so like -(-2)^2 + 2 = 6?

Answer 26

Not quite:
\[\lim_{x\to2}f(x)=\lim_{x\to2}(-x^2+2)=-2^2+2=-4+2=-2\](which agrees with the limit given by WA)

Answer 27

oh wow that was so easy, so if i i had put -1 it wouldn't work or number 1? 1 is closer to 2