- #1

ludi_srbin

- 137

- 0

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)

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- Thread starter ludi_srbin
- Start date

- #1

ludi_srbin

- 137

- 0

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)

- #2

Diane_

Homework Helper

- 393

- 1

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

ludi_srbin

- 137

- 0

So for others I just change ln to exp?

- #4

ludi_srbin

- 137

- 0

So I get f(x)^-1=e^(X^2-20)

and

f(x)^-1=e^x^2

and

f(x)^-1=e^x^2

- #5

Diane_

Homework Helper

- 393

- 1

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

ludi_srbin

- 137

- 0

It does.

Thank you so much.

Thank you so much.

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