Find the x-coordinates of all points on the graph of y=x^3 - 4x^2 + 5x at which the tangent line is horizontal.
Translate to math talk: find when dy/dx is zero.
y'=3x^2-8x+5=0
0=(3x-5)(x-1)
I assume you can do it from here.
Thanks
To find the x-coordinates of the points on the graph where the tangent line is horizontal, we need to find the points where the derivative of the function is equal to zero. When the derivative is zero, it indicates that the slope of the tangent line is horizontal, or in other words, the tangent line is flat.
Step 1: Find the derivative of the function.
To find the derivative of the function y = x^3 - 4x^2 + 5x, we differentiate each term separately using the power rule of differentiation. The power rule states that for a term of the form ax^n, the derivative is given by nx^(n-1).
So, differentiating each term of the function, we get:
dy/dx = 3x^2 - 8x + 5
Step 2: Set the derivative equal to zero and solve for x.
Setting the derivative equal to zero gives us the equation:
3x^2 - 8x + 5 = 0
This is a quadratic equation. To solve it, we can either factor it or use the quadratic formula.
Since the equation does not easily factor, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Using the values from the quadratic equation above, we have:
a = 3, b = -8, c = 5
Substituting these values into the quadratic formula, we get:
x = (-(-8) ± √((-8)^2 - 4(3)(5))) / (2(3))
x = (8 ± √(64 - 60)) / 6
x = (8 ± √4) / 6
x = (8 ± 2) / 6
Simplifying, we get two possible solutions:
x = (8 + 2) / 6 = 10 / 6 = 5/3
x = (8 - 2) / 6 = 6 / 6 = 1
Therefore, the x-coordinates of the points where the tangent line is horizontal are x = 5/3 and x = 1.