If the normal to f(x)= (3x/(1+x)) at (2,2) cuts the axes at B and C, determine the legnth of BC.
find the derivative of f(x)
find the value of that derivative at x = 2
call that the slope at (2,2)
then the normal has slope m = -1/derivative
find equation of that normal line, we know m already
y = m x + b
put in (2.2) to get b
Now all you have to do is find the x and y axis intercepts of that straight line.
f(x)= (3x/(1+x))
dy/dx = [(1+x)(3) - 3x ]/(1+x)^2
when x = 2
dy/dx = [9-6]/9 = 3/9 = 1/3
so the slope of the normal m = -3
You take it from there.
To determine the length of BC, we need to find the coordinates of points B and C.
Let's start by finding the equation of the normal to f(x) at (2,2).
The slope of the normal can be found by taking the negative reciprocal of the derivative of f(x) at x=2.
Derivative of f(x):
f'(x) = (d/dx) (3x/(1+x))
Using the quotient rule to differentiate the function, we get:
f'(x) = [(1+x)(3) - (3x)(1)] / (1+x)^2
= (3+3x - 3x) / (1+x)^2
= 3 / (1+x)^2
Now, let's find the slope of the normal at x=2.
f'(2) = 3 / (1+2)^2
= 3 / (9)
= 1/3
The slope of the normal is 1/3.
The equation of the normal can be written in the point-slope form:
y - y1 = m(x - x1)
Since the point (2,2) lies on the normal, we have:
y - 2 = (1/3)(x - 2)
Now, let's find the coordinates of points B and C.
Point C lies on the x-axis, so its y-coordinate is 0.
Substituting y=0 in the equation of the normal:
0 - 2 = (1/3)(x - 2)
-2 = (1/3)(x - 2)
Multiplying both sides by 3:
-6 = x - 2
x = -4
Thus, point C has coordinates (-4, 0).
Point B lies on the y-axis, so its x-coordinate is 0.
Substituting x=0 in the equation of the normal:
y - 2 = (1/3)(0 - 2)
y - 2 = (1/3)(-2)
y - 2 = -2/3
y = (2/3) + 2
y = 8/3
Thus, point B has coordinates (0, 8/3).
Now that we have coordinates of B and C, we can find the length of BC using the distance formula:
Length BC = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(0 - (-4))^2 + (8/3 - 0)^2]
= √[16 + (64/9)]
= √[(144 + 64)/9]
= √(208/9)
= √(208)/√(9)
= 4√(13)/3
Therefore, the length of BC is 4√(13)/3.
To determine the length of BC, we need to find the coordinates of points B and C.
To find the coordinates, we can utilize the given normal to the function f(x) at (2,2).
Step 1: Finding the slope of the normal line
The slope of the normal line can be obtained by taking the negative reciprocal of the derivative of f(x) at (2,2).
Step 2: Finding the equation of the normal line
Using the point-slope form of a line, we can write the equation of the normal line as y - y1 = m(x - x1), where (x1, y1) represents the given point (2,2), and m represents the slope of the normal line obtained in Step 1.
Step 3: Finding the coordinates of points B and C
The coordinates of B and C can be obtained by substituting the x-coordinates and y-coordinates of the respective axes into the equation of the normal line.
Step 4: Calculating the length of BC
After obtaining the coordinates of B and C, we can use the distance formula to calculate the length of BC, which is given by √((x2 - x1)^2 + (y2 - y1)^2).