posted by jane .
given f(x) = abs(sin x) and g(x) = x^2 for all real x:
1) let H(x) = g(f(x)) and write an expression for it.
2) find domain and range of H(x).
3) find an equation of the line tangent to the graph of H at the point where x = pi/4.
Since g(x) involves squaring x, you don't need the "absolute value" symbol when considering g(f(x)). You get the same result if you square a negative sinx. Thus
H(x) = g(f(x)) = sin^2 x
The domain is all real x and the range is from 0 to 1
At x = pi/4, H(x) = [sqrt2/2]^2 = 1/2
The slope dH/dx there is 2 sin x cos x = 1
So, find the equation of a line through x = pi/4, H = y = 1/2 with slope = 1
(y - 1/2)/(x - pi/4)= 1
y - 1/2 = x - pi/4
y = x + 1/2 - pi/4)
I am calling H(x) "y" in the graph of the function
i forgot the part that said "for -pi is less than or equal to x is less less than or equal to pi"
this was attached to the f(x) - does it change the answer? the domain?