Graphing e^f(x) and xf(x) (1 Viewer)

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Can any one explain to be general rules i should follow if i want to graph the following.
x f(x)
e^f(x)
cant seem to find it in any of my text books
 

Js^-1

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With x f(x) you should consider the following:

The origin is an x intercept, providid there are no points of discontinuity there.
If there are any horizontal asymptotes (not y=0) in the original function, then these become oblique asmpytotes in the new function with a gradient of the asymptotes y value. Eg. If the original function has a right hand horizontal asymptote at y=2, then the new function will have a right hand oblique asymptote to y = 2x. Similarly, if the original function is asymptotic to y=x, then the new function will be asymptotics to the parabola y=x<sup>2</sup>.
Any x-axis intercepts will be the same, plus the one at the origin.
At x=1, the graph of the new function will be intersect with the graph of the old function. At x= -1, the graph of the new function will be the reflection of the graph of the old function in the x-axis.

Basically, for positive x, the sign is the same as the old function, and for 0< x <1, then the new function will be smaller then the old function. For x>1 the function will be bigger.

For Negative x, the sign of the new function will be the opposite of the sign of the old function (a sign analysis is useful in the cases where there are many intercepts.) And the same rules apple for 0< x <1 and x > 1.


As for e<sup>f(x)</sup>, this is a little bit different, but not un-doable. We had one of these in our graphs assessment.

Basically, you look at the features of e<sup>x</sup>. When f(x) is 0, then e<sup>f(x)</sup> will be 1. If f(x) is ever negative, then e<sup>f(x)</sup> will be between 0 and 1. (Note, e<sup>f(x)</sup> is never equal to or less than 0. )
The larger the negative values of f(x), the closer the new graph gets to 0. If f(x) approaches negative infinity say, on the left hand side, then it will approach 0 on the left hand side.
If you have any horizontal asymptotes, then the value of e<sup>f(x)</sup> approaches e to the power of whatever the asymptote is. Eg. if the horizontal asymptote is to y=2, then the asymptote for e<sup>f(x)</sup> approaches e<sup>2</sup>.

Those are some of the main points I could think of, I'm sure there are more. Don't forget if they give you points, then you have to put those points in and show them on the new diagram. Eg, if (1,2) is on the original diagram, then (1,e<sup>2</sup>) will be on the new diagram.
Hope this helped...
 
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Js^-1

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No worries. Good luck with exams.
 

baller4life

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in addition to that query

when graphing y = ln f(X)

is the method just purely around the vertical assymptopes and the F(X) graph doesnt exist in the negative y?
 

Js^-1

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Um...kinda.

Basic properties should be included as well. Like when f(x)=1, the new graph will be 0.

If f(x) -> 0, then the new graph approaches negative infinity.

Other than that, just the no negative thing.

If you have a ln f(x) or e<sup>f(x)</sup> question, go across from left to right and input the function values of f(x) into the new function. You'll get a picture of what it looks like.
 

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