Find all points on the graph of f(x)=-x^3+3x^2-2 where there is a horizontal tangent line.
since it is a cubic, there will presumably be two
f' = 0 = 3 x^2 + 6 x = 3x ( x+2)
x = 0 and x = -2
you find f(x) there
Would the points be (0,-2) and -2, 18) when I plug them into f(x)?
Well, finding the points on the graph where there is a horizontal tangent line is like finding a needle in a haystack. Or better yet, a clown in a circus. We have to look for those special points where the slope of the graph is zero.
To find these points, we need to find the derivative of the function f(x). So, let's put on our clown shoes and differentiate this funky function!
The derivative of f(x)=-x^3+3x^2-2 can be found by applying the power rule. So, put your hands together and follow me:
f'(x) = -3x^2 + 6x
Now, we want to find the x-values where the derivative is equal to zero, because that's where we'll find our horizontal tangent line. So, let's set f'(x) equal to zero:
-3x^2 + 6x = 0
Now, let's simplify this equation. We can factor out an x:
x(-3x + 6) = 0
Now we have two possible solutions:
1) x = 0
2) -3x + 6 = 0
Let's solve the second equation:
-3x + 6 = 0
-3x = -6
x = 2
So, we have two values for x: x = 0 and x = 2.
Now, all that's left is to substitute these values back into our original function to find the corresponding y-values. And voila, we've found the points on the graph where there are horizontal tangent lines.
For x = 0, y = -0^3 + 3(0)^2 - 2 = -2.
For x = 2, y = -(2)^3 + 3(2)^2 - 2 = -6 + 12 - 2 = 4.
Therefore, the two points on the graph where there are horizontal tangent lines are (0, -2) and (2, 4).
So, there you have it! You can find your horizontal tangent line buddies at the points (0, -2) and (2, 4). Have fun clowning around on the graph!
To find the points on the graph of f(x) where there is a horizontal tangent line, we need to consider the derivative of the function f(x). A horizontal tangent line occurs when the derivative of the function is zero.
Step 1: Find the derivative of f(x):
f'(x) = -3x^2 + 6x
Step 2: Set f'(x) equal to zero and solve for x:
-3x^2 + 6x = 0
Factor out -3x:
-3x(x - 2) = 0
Setting each factor equal to zero:
-3x = 0 or x - 2 = 0
Solving each equation:
x = 0 or x = 2
So, there are two values of x that make the derivative zero, which means there are two points on the graph of f(x) where there is a horizontal tangent line.
Step 3: Find the corresponding y-values for the given x-values:
For x = 0:
f(0) = -(0)^3 + 3(0)^2 - 2
f(0) = 0 - 0 - 2
f(0) = -2
For x = 2:
f(2) = -(2)^3 + 3(2)^2 - 2
f(2) = -8 + 12 - 2
f(2) = 2
So, the two points on the graph of f(x) where there is a horizontal tangent line are (0, -2) and (2, 2).
f'=0=2x^2+6x
= 2x(x+3) so at x=0, x=-3, the tangent is horizontal.