Find all points at which the graph of f (x) = x^3 −3x has horizontal tangent lines.
f'(x) = 3x^2 - 3
= 0 for a horizontal tangent
3x^2 = 3
x^2 = 1
x = ± 1
if x=1, f(1)= 1-3 = -2
if x=-1 , f(-1) = -1+3 = 2
points are (1,-2) and (-1,2)
To find the points where the graph of f(x) = x^3 - 3x has horizontal tangent lines, we need to determine the values of x where the derivative of f(x) equals zero.
Let's denote the derivative of f(x) as f'(x). To find the derivative, we need to apply the power rule:
f'(x) = 3x^2 - 3
Now, set f'(x) = 0 and solve for x:
3x^2 - 3 = 0
Add 3 to both sides:
3x^2 = 3
Divide both sides by 3:
x^2 = 1
Taking the square root of both sides, we have:
x = ±1
Therefore, the values of x where the graph of f(x) has horizontal tangent lines are x = -1 and x = 1.
To find the corresponding points on the graph, substitute these values back into the original function:
For x = -1:
f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2
So, the point on the graph with a horizontal tangent line at x = -1 is (-1, 2).
For x = 1:
f(1) = (1)^3 - 3(1) = 1 - 3 = -2
Therefore, the point on the graph with a horizontal tangent line at x = 1 is (1, -2).
In summary, the graph of f(x) = x^3 - 3x has horizontal tangent lines at the points (-1, 2) and (1, -2).
To find the points at which the graph of a function has horizontal tangent lines, we need to find the critical points. Critical points occur where the derivative of the function is equal to zero or undefined.
Let's find the derivative of f(x) = x^3 - 3x first:
f'(x) = 3x^2 - 3
Now, since we want to find where the derivative is equal to zero, we set f'(x) = 0:
3x^2 - 3 = 0
To solve this equation, we can factor out a common factor:
3(x^2 - 1) = 0
Now, we set each factor equal to zero:
x^2 - 1 = 0
(x - 1)(x + 1) = 0
Now, we can solve for x:
x - 1 = 0 or x + 1 = 0
x = 1 or x = -1
So, the values of x are 1 and -1.
Now, let's find the corresponding y-values by substituting these x-values back into the original function:
f(1) = (1)^3 - 3(1) = 1 - 3 = -2
f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2
Therefore, the points at which the graph of f(x) = x^3 - 3x has horizontal tangent lines are (1, -2) and (-1, 2).