Let f(x)= √1-sinx)
(a.) What is the domain of x?
(b.) Write an equation for the tangent line to the graph of f at x=0.
I mostly just need help with letter B. I think I can get letter a.
f(x)=sqrt(1-sin(x))
Find derivative with respect to x (need to use chain rule):
f'(x) = -(1/2)cos(x)/sqrt(1-sin(x))
At x=0,
f(0)=1
f'(0)=-1/2 (=slope)
Therefore the line of the tangent has a slope of m=-1/2 and passes through (0,1)=(x0,y0).
The standard formula for a line with slope m passing through (x0,y0) is:
y=m(x-x0)+y0
Hope you can take it from here.
To find the equation for the tangent line to the graph of f at x=0, we will use the concept of the derivative. The derivative of f(x) can be found using the chain rule.
Step 1: Find f'(x) by applying the chain rule to f(x).
f(x) = √(1-sin(x))
f'(x) = (1/2)(1-sin(x))^(-1/2) * (-cos(x))
= -cos(x) / (2√(1-sin(x)))
Step 2: Evaluate f'(0) to find the slope of the tangent line at x=0.
f'(0) = -cos(0) / (2√(1-sin(0)))
= -1 / (2√(1-0))
= -1 / (2√1)
= -1/2
Step 3: Use the point-slope form of a line to find the equation of the tangent line.
The equation for the tangent line can be written as:
y - y1 = m(x - x1),
where m is the slope of the tangent line and (x1, y1) is a point on the line.
We already know that the slope of the tangent line is -1/2, and we are looking for the equation at x=0. Since x=0, we have a point (0, f(0)) on the line.
Step 4: Find the value of f(0) to determine the point (x1, y1).
To find f(0), substitute x=0 into the original function:
f(0) = √(1-sin(0))
= √(1-0)
= √1
= 1
So, the point (x1, y1) on the tangent line is (0, 1).
Step 5: Substitute the values into the point-slope form equation to get the final equation of the tangent line.
Using m = -1/2, x1 = 0, and y1 = 1, we have:
y - 1 = (-1/2)(x - 0)
y - 1 = (-1/2)x
Finally, we can simplify this equation to get the equation of the tangent line to the graph of f at x=0:
y = (-1/2)x + 1
To find the equation for the tangent line to the graph of f at x=0, we'll need to find the derivative of f(x) and then use it to determine the slope of the tangent line.
Step 1: Find the derivative of f(x)
The given function f(x) = √(1 - sinx) can be rewritten as f(x) = (1 - sinx)^(1/2). To find the derivative, we'll apply the chain rule.
f'(x) = (1/2)(1 - sinx)^(-1/2) * (-cosx)
= -(1/2)(cosx)/(√(1 - sinx))
Step 2: Find the slope of the tangent line
To find the slope of the tangent line at x=0, we substitute x=0 into the derivative:
f'(0) = -(1/2)(cos0)/(√(1 - sin0))
= -(1/2)(1)/(√(1 - 0))
= -1/2
Step 3: Find the equation of the tangent line
Now that we have the slope (-1/2) of the tangent line at x=0, we can use the point-slope form of a linear equation to find the equation of the tangent line. We'll use the point (0, f(0)) on the graph of f(x) to determine the y-intercept.
Given that f(x) = √(1 - sinx), plugging x=0 into the function, we find:
f(0) = √(1 - sin0)
= √(1 - 0)
= √1
= 1
So the point (0, f(0)) is (0, 1). Now we substitute the point (0, 1) and the slope (-1/2) into the point-slope form:
y - y1 = m(x - x1)
y - 1 = (-1/2)(x - 0)
y - 1 = (-1/2)x
y = (-1/2)x + 1
Therefore, the equation for the tangent line to the graph of f at x=0 is y = (-1/2)x + 1.