Find an equation of the line that is tangent to the graph of f and parallel to the given line.
Funtion:
f(x)=1/sqrt(x-1)
Line:
x+2y+7=0
f'=-1/2 * (x-1)^-1.5
that is the slope then tangent to the line.
y=-x/2-3.5
so slope is -1/2
f'=-1/2 * (x-1)^-3/2=-1/2
solving, x=2
then
y=1/sqrt(x-1)=1 so the point of tangency is 2,1
y=mx+b
2=-1/2 (1)+b
b=2.5
y=-x/2+2.5
check all this.
To find the equation of a line that is tangent to the graph of f and parallel to the given line, we need to determine the slope of both the graph of f and the given line.
First, let's find the slope of the given line. Rewriting the equation in slope-intercept form (y = mx + b), we have:
x + 2y + 7 = 0
2y = -x - 7
y = -1/2x - 7/2
The slope of this line is m = -1/2.
Next, we'll find the slope of the graph of f(x). The slope of a tangent line to a function is equal to the derivative of the function evaluated at the point of tangency.
f(x) = 1/sqrt(x - 1)
To find the derivative, we can use the power rule and chain rule.
f'(x) = -1/2(x - 1)^(-3/2)
Now, we need to find the value of x where the graph of f(x) is tangent to a line. Since the line is parallel to the given line, it will have the same slope of -1/2.
Set the derivative equal to -1/2:
-1/2(x - 1)^(-3/2) = -1/2
Simplifying, we have:
(x - 1)^(-3/2) = 1
To solve for x, we can raise both sides to the -2/3 power:
[(x - 1)^(-3/2)]^(-2/3) = 1^(-2/3)
(x - 1)^1 = 1
x - 1 = 1
x = 2
So, the tangent point is (2, f(2)). To find the y-coordinate, substitute x = 2 into the equation of f(x):
f(2) = 1/sqrt(2 - 1) = 1
Therefore, the point of tangency is (2, 1).
Now, we have the slope of the tangent line (m = -1/2) and the point of tangency (2, 1). Using the point-slope form of the equation of a line, we can write the equation:
y - y1 = m(x - x1)
y - 1 = (-1/2)(x - 2)
Simplifying further, we get:
y - 1 = -1/2x + 1
y = -1/2x + 2
Therefore, the equation of the line that is tangent to the graph of f and parallel to the given line is y = -1/2x + 2.