We are given the function to name as f(x) equal to 3x + 12 and we are asked to find the inverse of the function and name it g(x). first step is to extract x, this is equal to (y - 12)/ 3 = x; then exchange the places of x and y, that is (x- 12)/3 = y = g(x). we evaluate f(g(-2)). f(g(x)) is equal to 3*((x- 12)/3) + 12 = x - 12 + 12 = x. The domain is from negative infinity to positive infinity
we are given the equation sin^2(x)cos^2(x) = 1/4 and is asked to evaluate the x-value of the equation given. we first start by taking the square root of both sides resulting to sin (x) cos(x) = 1/2. By double angle identity, 1/2 sin 2x = 1/2. Simplifying, sin 2x = 1; x is equal to pi/4.
write 1/(1-y2) into form -1/ (y^2- 1) = -1/ ( ( y+1) (y-1) );
then use partial fraction
let -1/ ( ( y+1) (y-1) ) = A/ (y+1) + B/ (y-1);
multiplying the whole equation by ( y+1) (y-1)
therefore A(y-1) + B (y+1) = -1
y(A +B) +B - A = -1
equating the constants:
B - A = -1 equating y: A + B = 0
adding these two 2B =1 , B =1/2
A= 1/2B= -1/2
then [ -1/ ( y^2-1) ] = 1/2 ( 1/ (y+1) ) - 1/2 ( 1/ (y -1) )
The integral is equal to 1/2 ln (y+1) -1/2 ln (y-1)
