Find the equation of each tangent line to the curve y=(1/x) that has a slope of -1. Find each point of tangency.
if y = 1/x
then y' = -1/x^2
given , y' = -1
-1/x^2= - 1
x^2 = 1
x = ± 1
find the corresponding y values.
You have 2 tangents, one at (1,1) , the other at (-1,-1)
Use your favourite method to find the equation of those tangents.
(The question is poorly worded when it asked for the equation of the tangent, implying there is only one)
Why did the tangent line go to the circus? Because it wanted to find the curve that y=(1/x) and have a slope of -1!
To find the equation of the tangent line, we need to first find the derivative of y=(1/x). The derivative will give us the slope of the curve at each point.
Using the power rule, we find that the derivative of y=(1/x) is y' = -1/x^2.
Since we want the slope to be -1, we can set y' equal to -1 and solve for x:
-1/x^2 = -1
Dividing both sides by -1, we get:
1/x^2 = 1
Taking the square root of both sides, we have:
1/x = 1
Multiplying both sides by x, we get:
x = 1
Now we have the x-coordinate of the point of tangency. To find the corresponding y-coordinate, we substitute x=1 back into the original equation y=(1/x):
y = 1/1 = 1
So the first point of tangency is (1, 1).
Now we can find the equation of the tangent line. We have the point of tangency (1, 1) and we know the slope is -1. We can use the point-slope form of a line to find the equation:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is the point of tangency.
Substituting the values, we get:
y - 1 = -1(x - 1)
Simplifying, we have:
y - 1 = -x + 1
Adding x to both sides, we get:
y + x - 1 = 1
Rearranging, we find:
y + x = 2
So the equation of the first tangent line is y + x = 2.
To find the second point of tangency, we can repeat the steps:
-1/x^2 = -1
Dividing both sides by -1, we get:
1/x^2 = 1
Taking the square root of both sides, we have:
1/x = 1
Multiplying both sides by x, we get:
x = 1
Now we substitute x=1 back into the original equation y=(1/x):
y = 1/1 = 1
So the second point of tangency is (1, 1).
Using the point-slope form, we can find the equation of the tangent line:
y - 1 = -1(x - 1)
Simplifying, we have:
y - 1 = -x + 1
Adding x to both sides, we get:
y + x - 1 = 1
Rearranging, we find:
y + x = 2
So the equation of the second tangent line is y + x = 2.
Therefore, the equations of both tangent lines are y + x = 2, and the points of tangency are (1, 1) for both lines.
Looks like our tangent lines couldn't get enough of the circus!
To find the equation of each tangent line to the curve y=(1/x) with a slope of -1, we need to find the x-values at which the slope of the curve is -1 and then find the corresponding y-values.
Step 1: Find the derivative of the curve y=(1/x) using the quotient rule. The derivative will give us the slope of the curve at any point.
dy/dx = -1/x^2
Step 2: Set the derivative equal to -1 to find the x-values where the slope is -1.
-1/x^2 = -1
Solving this equation, we get: x^2 = 1
Taking the square root, we have two possible solutions: x = 1 and x = -1.
Step 3: Find the corresponding y-values by substituting the x-values into the equation y = 1/x.
For x = 1, y = 1/1 = 1.
For x = -1, y = 1/-1 = -1.
So the two points of tangency are (1, 1) and (-1, -1).
Step 4: Find the equations of the tangent lines using the point-slope form.
For the point (1, 1):
Using the point-slope form: y - y1 = m(x - x1)
We have: y - 1 = -1(x - 1)
Simplifying, we get: y - 1 = -x + 1
Rearranging the equation, we have: y = -x + 2
For the point (-1, -1):
Using the point-slope form: y - y1 = m(x - x1)
We have: y - (-1) = -1(x - (-1))
Simplifying, we get: y + 1 = -x - 1
Rearranging the equation, we have: y = -x - 2
Therefore, the equation of the tangent line to the curve y=(1/x) with a slope of -1 at the point (1, 1) is y = -x + 2, and the equation of the tangent line at the point (-1, -1) is y = -x - 2.