an equation of the normal to the graph of f(x)= x/(2x-3) at (1,f(1)
Why did the normal to the graph of f(x) go to therapy?
Because it had a lot of issues with its slope! But don't worry, I can help you find the equation.
To find the equation of the normal to the graph, we need to find its slope. The slope of the graph of f(x) at any given point can be found by taking the derivative of f(x).
So, let's find the derivative of f(x):
f'(x) = (2 - 3x) / (2x - 3)^2
Now, to find the slope of the normal at (1, f(1)), we substitute x = 1 into f'(x):
f'(1) = (2 - 3(1)) / (2(1) - 3)^2
= -1 / 1
= -1
So the slope of the normal is -1.
Now we can use the point-slope form of a line to find the equation of the normal:
y - f(1) = -1 * (x - 1)
Simplifying this equation gives us the equation of the normal to the graph of f(x) at (1, f(1)).
To find the equation of the normal to the graph of f(x) at a specific point, you need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope to get the slope of the normal.
Let's start by finding the slope of the tangent line at the point (1, f(1)).
1. Find the derivative of f(x) with respect to x:
f'(x) = d/dx [x/(2x-3)]
To find the derivative of f(x), we can use the quotient rule:
f'(x) = [(2x-3)(1) - (x)(2)] / (2x-3)^2
Simplifying the expression:
f'(x) = (2x - 3 - 2x) / (2x - 3)^2
f'(x) = -3 / (2x - 3)^2
2. Substitute x = 1 into f'(x) to find the slope at x = 1:
f'(1) = -3 / (2(1) - 3)^2
f'(1) = -3 / (-1)^2
f'(1) = -3
So, the slope of the tangent line at (1, f(1)) is -3.
Now, let's find the slope of the normal line:
The slope of the normal line is the negative reciprocal of the slope of the tangent line.
Therefore, the slope of the normal line = -1 / (-3) = 1 / 3.
Now, we have the slope of the normal line, which is 1/3, and we know that it passes through (1, f(1)).
Using the point-slope form of a linear equation, we can calculate the equation of the normal line:
y - y1 = m(x - x1)
y - f(1) = (1/3)(x - 1)
Substituting the values:
y - f(1) = (1/3)(x - 1)
Therefore, the equation of the normal to the graph of f(x) = x/(2x-3) at (1, f(1)) is y - f(1) = (1/3)(x - 1).
To find the equation of the normal to the graph of the function f(x) at a specific point (1, f(1)), you need to follow these steps:
Step 1: Find the derivative of the function f(x).
Step 2: Substitute x = 1 in the derivative to find the slope of the tangent line at the point (1, f(1)).
Step 3: Find the negative reciprocal of the slope from step 2 to get the slope of the normal line.
Step 4: Use the point-slope form of a line with slope from step 3 and the point (1, f(1)) to find the equation of the normal line.
Let's go through the steps one by one.
Step 1: Find the derivative of f(x).
To find the derivative of f(x), you can use the quotient rule. The quotient rule states that if you have a function in the form of u(x)/v(x), its derivative is given by (v(x)*u'(x) - u(x)*v'(x))/[v(x)]^2.
In this case, u(x) = x and v(x) = 2x - 3. Thus, differentiating f(x) gives us:
f'(x) = [(2x - 3)(1) - (x)(2)] / [(2x - 3)^2]
= (2x - 3 - 2x) / [(2x - 3)^2]
= -3 / [(2x - 3)^2]
Step 2: Substitute x = 1 in the derivative.
f'(1) = -3 / [(2(1) - 3)^2]
= -3 / [(2 - 3)^2]
= -3 / (-1)^2
= -3 / 1
= -3
Step 3: Find the negative reciprocal of the derivative.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is 1/3.
Step 4: Use the point-slope form of a line.
The point-slope form of a line is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Using the point (1, f(1)) = (1, 1/(-1)) = (1, -1), we have:
y - (-1) = (1/3)(x - 1)
y + 1 = (1/3)(x - 1)
This is the equation of the normal line to the graph of f(x) at the point (1, f(1)).
find the slope of f(x), or
f'= d/dx (f(x)) at the point.
Then, the slope to the normal is the negative reciprocal of that value.