Find all points on the curve where the tangent line is horizontal
y = (x^6/6) - (x^2/2) + 1
Well, finding where the tangent line is horizontal means finding where the derivative of the function is equal to zero. Let's take the derivative of the function and set it equal to zero to find those points.
To find the derivative of y with respect to x, we can apply the power rule and get:
dy/dx = (6/6)x^5 - (2/2)x^1 + 0
Simplifying that, we have:
dy/dx = x^5 - x
Now, let's set dy/dx equal to zero and find the values of x that satisfy the equation:
x^5 - x = 0
Factoring out an x, we get:
x(x^4 - 1) = 0
Setting each factor equal to zero:
x = 0 or x^4 - 1 = 0
From the first factor, we get x = 0.
For the second factor, x^4 - 1 = 0, we can factor it as a difference of squares:
(x^2 - 1)(x^2 + 1) = 0
Setting each factor equal to zero:
x^2 - 1 = 0 or x^2 + 1 = 0
For x^2 - 1 = 0, we find x = -1 and x = 1.
For x^2 + 1 = 0, there are no real solutions.
So, the points where the tangent line is horizontal on the curve are x = 0, x = -1, and x = 1.
To find the corresponding y-values, substitute these x-values back into the original equation:
When x = 0:
y = 0^6/6 - 0^2/2 + 1 = 1
When x = -1:
y = (-1)^6/6 - (-1)^2/2 + 1 = 1/6
When x = 1:
y = 1^6/6 - 1^2/2 + 1 = 7/6
So, the points on the curve where the tangent line is horizontal are (0, 1), (-1, 1/6), and (1, 7/6).
I hope this answer didn't drive you too round the bend!
To find the points on the curve where the tangent line is horizontal, we need to find the values of x where the derivative of y with respect to x is equal to 0.
Step 1: Find the derivative of y with respect to x.
To find the derivative of y, we need to use the power rule:
dy/dx = d/dx [(x^6/6) - (x^2/2) + 1]
= (6x^5/6) - (2x/2)
= x^5 - x
Step 2: Set the derivative equal to 0 and solve for x.
We set dy/dx = 0 and solve for x:
x^5 - x = 0
Step 3: Factor out x from the equation.
x(x^4 - 1) = 0
Step 4: Set each factor equal to 0 and solve for x.
First, we consider x = 0:
x = 0
Next, we consider x^4 - 1 = 0:
x^4 - 1 = 0
(x^2 + 1)(x^2 - 1) = 0
(x^2 + 1)(x + 1)(x - 1) = 0
So, x can be either -1, 0, or 1.
Step 5: The x-values we found represent the points on the curve where the tangent lines are horizontal. Now, we substitute these values of x into the original equation to find the corresponding y-values.
When x = -1:
y = (-1^6/6) - (-1^2/2) + 1
= (1/6) - (1/2) + 1
= 1/6 - 3/6 + 6/6
= 4/6
= 2/3
So, one point on the curve where the tangent line is horizontal is (-1, 2/3).
When x = 0:
y = (0^6/6) - (0^2/2) + 1
= 0 - 0 + 1
= 1
So, another point on the curve where the tangent line is horizontal is (0, 1).
When x = 1:
y = (1^6/6) - (1^2/2) + 1
= (1/6) - (1/2) + 1
= 1/6 - 3/6 + 6/6
= 4/6
= 2/3
So, the last point on the curve where the tangent line is horizontal is (1, 2/3).
Therefore, the points on the curve where the tangent line is horizontal are (-1, 2/3), (0, 1), and (1, 2/3).
To find all points on the curve where the tangent line is horizontal, we need to find where the derivative of the function is equal to zero. The derivative represents the slope of the tangent line at each point on the curve.
Let's find the derivative of the given function first:
y = (x^6/6) - (x^2/2) + 1
To differentiate each term, we can use the power rule:
(d/dx) (x^n) = n*x^(n-1)
Using the power rule, we can find the derivative:
(dy/dx) = (1/6)*(6*x^(6-1)) - (1/2)*(2*x^(2-1))
Simplifying the expression gives us:
dy/dx = x^5 - x
Now, to find where the tangent line is horizontal, we set dy/dx equal to zero and solve for x:
0 = x^5 - x
To solve this equation, we can factor out x:
0 = x(x^4 - 1)
Setting each factor equal to zero:
x = 0,
x^4 - 1 = 0
For the first factor, x = 0.
For the second factor, x^4 - 1 = 0, we can rearrange the equation to find:
x^4 = 1
Taking the fourth root of both sides gives us:
x = ±1
Therefore, the points on the curve where the tangent line is horizontal are (0,1), (-1, -1), and (1, -1).