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Could you check this please

  • Thread starter ludi_srbin
  • Start date
  • #1
137
0
f(x)=e^cosx

equation for the axis of symmetry???
I got x=pi


f(x)=ln(x^2 - 20)

describe symmetry??

I got symmetric arround y axis.

inverse??

I got y=sqr of ((x/ln) +20)

f(x)=ln(x^2)

Inverse???

I got y= sqr of (x/ln)
 

Answers and Replies

  • #2
Diane_
Homework Helper
390
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Since cos x is a repeating function, it has an infinite number of axes of symmetry. They occur at n pi. Something similar needs to be done for exp(cos x).

For the others, you seem to be using "ln" as a variable of some sort. Remember that ln x is a function, with inverse exp(x). You might want to look at those again.
 
  • #3
137
0
So for others I just change ln to exp?
 
  • #4
137
0
So I get f(x)^-1=e^(X^2-20)

and

f(x)^-1=e^x^2
 
  • #5
Diane_
Homework Helper
390
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Let me take one of them. You should be able to do the rest.

y = ln(x^2 - 20)

exp(y) = x^2 - 20

x^2 = exp(y) + 20

x = sqrt(exp(y) + 20)

So, f^-1(x) = sqrt(exp(x) + 20)

Note that there are restrictions on the domain and range - these would correspond to the restrictions on the range and domain of the original function.

Does that help?
 
  • #6
137
0
It does.

Thank you so much. :smile:
 

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