if f(x)=sinx on 0<(or equal to) to x<(or equal to) 2(pie)
and h(x)=ln x on it's domain, state the domain of h(f(x)
Given domain of f(x)=sin(x) is [0,2π].
The range of f(x) is [-1,1].
Domain of h(x)=log(x) is ]0,&#infin[.
The domain of h(f(x)) is equivalent to that of h(range of f(x))
Since f(0)=0, f(π)=0 and f([π,2π])≤0, they are outside the domain of h(x)=log(x), they must be excluded from the domain of h(f(x)).
Thus the domain of h(f(x)) is
{x ∈ ℝ | ]0,π[}
erratum:
Domain of h(x)=log(x) is ]0,&#infin;[.
erratum:
I'll try it again:
Domain of h(x)=log(x) is ]0,∞[.
To find the domain of the composite function h(f(x)), we need to consider two things:
1. The domain of f(x) which is given as 0 ≤ x ≤ 2π.
2. The domain of h(x), which is determined by the input values for the natural logarithm function.
The natural logarithm function, ln(x), is defined only for positive real numbers. So, for h(x) = ln(x) to be defined, the input (x) must be greater than 0.
Now, let's consider the composition h(f(x)):
h(f(x)) = h(sin(x))
Since sin(x) can take any value between -1 and 1 for any real number x, we need to determine the range of sin(x) within the given domain 0 ≤ x ≤ 2π.
The range of sin(x) on this interval is [-1, 1]. Therefore, h(f(x)) = h(sin(x)) will be defined as long as sin(x) is positive, i.e., sin(x) > 0.
In the given domain of 0 ≤ x ≤ 2π, sin(x) is positive in the interval (0, π). Thus, the domain of h(f(x)) is the interval (0, π).