INVERSE HYPERBOLIC FUNCTIONS

We will recall that the function eex − − x sinh x = 2 is a 1 to 1 mapping.ie one value of x maps onto one value of sinhx.

This implies that the inverse function exists.

y =⇔=sinhxx sinh−1 y The function f(x)=sinhx clearly involves the exponential function.

We will now find the inverse function in its logarithmic form. LET y = sinh−1 x x = sinh y THEN eey − − y x = 2 (BY DEFINITION) SOME ALGEBRAIC MANIPULATION GIVES

yy− 2xe=− e

MULTIPLYING BOTH SIDES BY e y

yy2 21xe= e −

RE-ARRANGING TO CREATE A QUADRATIC IN e y 2 yy 02=−exe −1 APPLYING THE QUADRATIC FORMULA

24xx± 2 + 4 e y = 2 22xx± 2 + 1 e y = 2

y 2 exx= ±+1 SINCE y e >0 WE MUST ONLY CONSIDER THE ADDITION.

y 2 exx= ++1 TAKING LOGARITHMS OF BOTH SIDES

yxx=++ln2 1 { }

IN CONCLUSION

−12 sinhxxx=++ ln 1 { } THIS IS CALLED THE LOGARITHMIC FORM OF THE INVERSE FUNCTION.

The graph of the inverse function is of course a reflection in the line y=x.

THE INVERSE OF COSHX

The function eex + − x fx()== cosh x 2 Is of course not a one to one function. (observe the graph of the function earlier). In order to have an inverse the DOMAIN of the function is restricted.

eexx+ − fx()== cosh x , x≥ 0 2 This is a one to one mapping with RANGE [1, ∞] .

We find the inverse function in logarithmic form in a similar way.

y = cosh −1 x x = cosh y provided y ≥ 0 eeyy+ − xy==cosh 2 2xe=+yy e− 02=−+exeyy− 2 yy 02=−exe +1

USING QUADRATIC FORMULA

−24xx±−2 4 e y = 2

y 2 exx=± −1 TAKING LOGS yxx=±−ln2 1 { } yxx=+−ln2 1 is one possible expression { } APPROPRIATE IF THE DOMAIN OF THE ORIGINAL FUNCTION HAD BEEN RESTRICTED AS EXPLAINED ABOVE.

HOWEVER CONSIDERING THE WHOLE FUNCTION (UNRESTRICTED) FOR EACH VALUE OF X THERE ARE TWO EQUAL AND OPPOSITE VALUES OF Y.

This implies cosh−12xxx= ±+− ln 1 { } To add a more sound mathematical argument to this we can use some algebraic manipulation and laws of logs to the alternative solution.

Consider (1)(xx− 22−+− xx1) xx−−≡2 1 (1)xx+−2 THIS IS NOTHING BUT A USEFUL TRICK!

The numerator becomes 1 (check it and see!)

So 1 xx−−≡221( ≡+−xx1)−1 (1)xx+−2

Now returning to out alternative solution

−12 yxxx==±+cosh ln− 1 { }

Covers all possible solutions depending on the restriction of the domain TASK: Show that the logarithmic form of the hyperbolic tan is

−1 1 1+ x y ==tanhxx ln for ∠1 HINT 2 1− x Start as before by rearranging the inverse statement and then use eex − −x tanh x = eex + −x

THE GOOD NEWS! THE FORMULAE FOR THE LOGARITHMIC FORMS OF INVERSE HYPERBOLIC FUNCTIONS ARE IN THE WJEC FORMULA BOOK!