Find the equation of...
a) the tangent line
b) the normal line
of the equation: f(x) = √x^2 - 1, @ x=1
Show the correct solution
AAAaannndd the bot gets it wrong yet again!
see your annoyingly repeated post above somewhere.
To find the equation of the tangent line and the normal line of the function f(x) = √(x^2 - 1) at x = 1, we need to follow these steps:
Step 1: Find the slope of the tangent line
We can find the slope of the tangent line by taking the derivative of the function f(x). Let's start by finding the derivative of f(x).
f(x) = √(x^2 - 1)
f'(x) = (1/2)(x^2 - 1)^(-1/2) * 2x
Now, let's substitute x = 1 into f'(x) to find the slope of the tangent line.
f'(1) = (1/2)(1^2 - 1)^(-1/2) * 2(1)
f'(1) = (1/2)(0)^(-1/2) * 2
f'(1) = undefined
We see that f'(1) is undefined, which means the slope of the tangent line at x = 1 does not exist.
Since we cannot find the slope of the tangent line, we cannot proceed to find its equation.
However, we can still find the equation of the normal line.
Step 2: Find the equation of the normal line
The normal line is a line that is perpendicular to the tangent line. Since the slope of the tangent line is undefined, the slope of the normal line will be 0.
Now let's find the equation of the normal line passing through the point (1, f(1)).
The point (1, f(1)) is (1, √(1^2 - 1)) = (1, 0).
The equation of a line with slope m passing through the point (x1, y1) is given by the formula: y - y1 = m(x - x1).
Since the slope of the normal line is 0, the equation of the normal line passing through (1, 0) will be y - 0 = 0(x - 1) or simply y = 0.
Therefore, the equation of the normal line is y = 0.
To find the equation of the tangent line and normal line of a function at a specific point, you can follow these steps:
Step 1: Find the derivative of the function.
Step 2: Substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line.
Step 3: Use the point-slope form to determine the equation of the tangent line.
Step 4: Use the negative reciprocal of the slope of the tangent line to find the slope of the normal line.
Step 5: Use the point-slope form again to determine the equation of the normal line.
Now let's apply these steps to the function f(x) = √x^2 - 1 and find the equation of the tangent line and normal line at x = 1.
a) The equation of the tangent line:
Step 1: Find the derivative of f(x):
f'(x) = (1/2)*(2x) / √x^2 - 1 (using the Quotient Rule)
= x / √x^2 - 1
Step 2: Substitute x = 1 into f'(x) to find the slope of the tangent line at x = 1:
f'(1) = 1 / √1^2 - 1
= 1 / √1 - 1
= 1 / 0 (since √1 - 1 = 0)
= Undefined
Since the slope is undefined, it means the tangent line is vertical at x = 1. The equation of the tangent line will be in the form x = a, where a is the x-coordinate of the point of tangency.
Therefore, the equation of the tangent line is x = 1.
b) The equation of the normal line:
Step 4: Find the slope of the normal line. Since the slope of the tangent line is undefined (vertical), the slope of the normal line will be 0.
Step 5: Use the point-slope form using the point of tangency (1, f(1)):
y - f(1) = 0 * (x - 1)
y - f(1) = 0
y = f(1)
Substitute the x-coordinate into the original function:
y = √1^2 - 1
= √1 - 1
= 0
Therefore, the equation of the normal line is y = 0.
a) The equation of the tangent line is y = 2x - 1.
b) The equation of the normal line is y = -1/2x + 1.