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?
To solve these questions, we need to follow a set of steps. Let's go through them one by one:
1) To find H(x), we substitute the function f(x) into g(x). So, H(x) = g(f(x)). Plugging in the given expressions for f(x) and g(x), we have:
H(x) = g(f(x)) = (f(x))^2 = (abs(sin x))^2 = (sin x)^2.
2) To determine the domain and range of H(x), we consider the domain and range of f(x) and g(x).
The function f(x) = abs(sin x) has a domain of all real numbers since sin x is defined for all x, and the absolute value function also applies to all real numbers.
The function g(x) = x^2 is defined for all real numbers, so its domain is also all real numbers.
Therefore, the domain of H(x) is the same as the domain of f(x) and g(x), which is all real numbers.
For the range of H(x), the function (sin x)^2 is always greater than or equal to 0, since squaring any real number results in a non-negative value. Thus, the range of H(x) is all non-negative real numbers or [0, +∞).
3) To find the equation of the tangent line to the graph of H(x) at x = π/4, we need to determine its slope and the point of tangency.
To find the slope, we can differentiate H(x) = (sin x)^2 with respect to x:
H'(x) = 2sin x * cos x.
Plugging in x = π/4, we have:
H'(π/4) = 2sin(π/4) * cos(π/4) = 2(1/sqrt(2)) * (1/sqrt(2)) = 1/2.
So, the slope of the tangent line at x = π/4 is 1/2.
To find the point of tangency, we evaluate H(π/4):
H(π/4) = (sin(π/4))^2 = (1/sqrt(2))^2 = 1/2.
Therefore, the point of tangency is (π/4, 1/2).
Using the point-slope form of a line, the equation of the tangent line with slope 1/2 and passing through (π/4, 1/2) is:
y - 1/2 = (1/2)(x - π/4).
Expanding this equation, we have:
y = (1/2)x - 1/8 + 1/2.
Simplifying, we get:
y = (1/2)x + 3/8.
Thus, the equation of the tangent line to the graph of H at x = π/4 is y = (1/2)x + 3/8.