Find the points on y=(-x^2+4x-3)^3 that has a horizontal tangent line.
y' = 3(-x^2+4x-3)^2 (-2x+4)
y'=0 when x = 1,2,3
so, the tangent is horizontal at
(1,0),(2,1),(3,0)
Tnx! :D
To find the points on the curve y = (-x^2 + 4x - 3)^3 that have a horizontal tangent line, we need to find the derivative of the function and set it equal to zero.
Step 1: Find the derivative of the function y = (-x^2 + 4x - 3)^3 with respect to x.
To find the derivative, we can use the chain rule. Let's denote the function inside the parentheses as u(x) = -x^2 + 4x - 3. Applying the chain rule, the derivative of y with respect to x can be found as:
dy/dx = d(u(x)^3)/dx = 3(u(x)^2) * du(x)/dx.
Step 2: Set the derivative equal to zero to find where it has a horizontal tangent line.
Since we want to find the points where the tangent line is horizontal, we need to solve the equation dy/dx = 0. Therefore, we have:
3(u(x)^2) * du(x)/dx = 0.
Step 3: Solve the equation to find the value(s) of x that yield a horizontal tangent line.
To find the values of x, we need to solve the equation 3(u(x)^2) * du(x)/dx = 0. This equation can be satisfied in two cases:
Case 1: 3(u(x)^2) = 0. In this case, u(x) = 0.
We can solve u(x) = -x^2 + 4x - 3 = 0 by factoring or using the quadratic formula. However, since this equation is not a quadratic, factoring may lead to a complex solution. Therefore, let's use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a).
Applying this to our equation u(x) = -x^2 + 4x - 3 = 0, we obtain:
x = [-4 ± √(4^2 - 4(-1)(-3))] / (2(-1)).
Simplifying further:
x = [-4 ± √(16 - 12)] / (-2).
x = [-4 ± √4] / (-2).
x = [-4 ± 2] / (-2).
x = -2 ± 1.
So we have two solutions for x: x1 = -3 and x2 = -1.
Case 2: du(x)/dx = 0.
To find du(x)/dx, we take the derivative of u(x) = -x^2 + 4x - 3 again:
du(x)/dx = -2x + 4.
We set this derivative equal to zero:
-2x + 4 = 0.
2x = 4.
x = 2.
Therefore, x = 2 is another solution.
Step 4: Find the corresponding y-values for the points where the tangent line is horizontal.
To find the corresponding y-values, we substitute the values of x we found (x1 = -3, x2 = -1, and x = 2) into the original equation y = (-x^2 + 4x - 3)^3:
For x1 = -3:
y1 = (-(-3)^2 + 4(-3) - 3)^3
= (9 - 12 - 3)^3
= (-6)^3
= -216.
For x2 = -1:
y2 = (-(-1)^2 + 4(-1) - 3)^3
= (1 - 4 - 3)^3
= (-6)^3
= -216.
For x = 2:
y = (-(2)^2 + 4(2) - 3)^3
= (4 + 8 - 3)^3
= (9)^3
= 729.
So the points on the curve y = (-x^2 + 4x - 3)^3 that have a horizontal tangent line are (-3, -216), (-1, -216), and (2, 729).