Question

the domain of function f(x)=sec-1(3x-4)+tan-1(x+3/5)is​

Answer 1

Answer:

The domain of function f(x)= sec-1(3x-4)+tan-1(x+3/5)is

Answer 2

Hence, Domain of f(x) = $( - 8,1] U [ \frac{5}{3},2)$

GIVEN :- f(x)= {sec}^{ - 1} (3x - 4) + {tanh}^{ - 1} ( \frac{x + 3}{5} )[/tex]

TO FIND :- Domain function

SOLUTION:-

The function f(x) =

${sec}^{ - 1} (3x - 4) + {tanh}^{ - 1} ( \frac{x + 3}{5} )$

$= {sec}^{ - 1} (3x - 4) + \frac{1}{2} ln \: \frac{1 + \frac{x +3 }{5} }{1 - \frac{x - 3}{5} }$

$= {sec}^{ - 1} (3x - 4) + \frac{1}{2} ln( \frac{8 + x}{2 - x})$

In

${sec}^{ - 1}( 3x - 4)$

$= 3x - 4 \leqslant - 1 \: U \: 3x - 4 \geqslant 1$

$= 3x \leqslant 3 \: U \: 3x \geqslant 5$

$= x \leqslant 1U x \geqslant \frac{5}{3}$

In

$ln( \frac{8 + x}{2 - x} )$

$\frac{8 + x}{2 - x} > 0$

Hence, Domain of f(x) =

$( - 8,1] U [ \frac{5}{3},2)$