Determine the points on the curve y=x^2(x^3-x)^2 where the tangent line is horizontal. Thx!

(y z x2)

The slope of the tangent line is equal to the first derivative.

You must find where is first derivative=0

In this case the tangent line is horizontal.

In google type: calc101

When you see list of results click on:

Calc101com Automatic Calculus,Linear Algebra and Polynomials

When page be open clik option: derivatives

When this page be open in rectacangle type your equation and click options
DO IT

You will see solution for first derivative.

Then in google type: equation solver

When you see list of results click on:

Equation Calculator & Solver-Algebra.help

When this page be open in rectacangle type expresion for firste derivatives=0 and click options solve.

If your expresion is:
x^2*(x^3-x)^2

First derivative is:
4(2x^7-3x^5+x^3)

In this case in equation solver type:
4(2x^7-3x^5+x^3)=0

and solutions is:

-1, -sqroot(2)/2, sqroot(2)/2, 1

One of solutions is obviously:

x=0

To find the points on the curve where the tangent line is horizontal, we need to determine the values of x where the slope of the tangent line is zero.

First, let's find the derivative of the curve y = x^2(x^3 - x)^2 with respect to x.

Take the product rule to differentiate the expression:

y = x^2 (x^3 - x)^2

Let u = x^2 and v = (x^3 - x)^2. Then, using the chain rule, the derivative dy/dx can be written as:

dy/dx = u * d(v)/dx + v * d(u)/dx

Now let's find d(v)/dx and d(u)/dx:

d(u)/dx = 2x
d(v)/dx = 2(x^3 - x) * (3x^2 - 1)

Plugging these values into the formula, we get:

dy/dx = x^2 * 2(x^3 - x)(3x^2 - 1) + (x^3 - x)^2 * 2x

Simplifying further, we have:

dy/dx = 2x(x^3 - x)(3x^2 - 1) + 2x(x^3 - x)^2

To find the values of x where the tangent line is horizontal, we need to solve the equation dy/dx = 0:

2x(x^3 - x)(3x^2 - 1) + 2x(x^3 - x)^2 = 0

Now let's solve this equation to find the values of x.