The slope of the tangent line to the graph y= x^3/x -x at the point (1, -2/3)

I keep getting undefined as a answer but.. I feel like I might be doing something wrong because I feel like that is not the answer, can you help me please?

could be the way you typed your equation.

The way you typed it, why not just simplify x^3/x to x^2 ?
That is why I am suspicious about the way it was typed.
Did you mean y = x^(3/x) - x ?

The equation is y= x cubed over x minus x

so;

y= x^3 / x -x

Sorry for being confusingg

y = x^3/x- x would reduce to

y = x^2 - x
and (1,-2/3) would NOT lie on this curve.

So there is something wrong with the way you typed it

Use brackets to establish the correct form
Otherwise this question is bogus.

Okay I sure will! Thanks

Hmmm you know what, I think this problem is an error. Maybe that's why I could not figure it outt?

Sure, I can help you. To find the slope of the tangent line to the graph of the function at a specific point, you need to take the derivative of the function and then substitute the x-coordinate of the given point into the derivative.

Let's start by finding the derivative of the function y = x^3/x - x. To do that, we can use the quotient rule, which states that if you have a function of the form f(x) = g(x)/h(x), where both g(x) and h(x) are differentiable functions, then the derivative of f(x) is given by (h(x)*g'(x) - g(x)*h'(x))/(h(x))^2.

Let's apply the quotient rule to our function y = x^3/x - x:
g(x) = x^3
h(x) = x

Using the quotient rule, we have:
y' = (x*h'(x) - h(x)*g'(x))/(h(x))^2

Now, let's find the derivatives of g(x) and h(x):
g'(x) = 3x^2
h'(x) = 1

Substituting these derivatives into the quotient rule, we get:
y' = (x*1 - x^3*1)/(x^2)^2
= (x - x^3)/(x^4)
= (1 - x^2)/x^3

Now, we can substitute x = 1 into y' to find the slope at the point (1, -2/3):
slope = (1 - (1)^2)/(1)^3
= (1 - 1)/1
= 0

Therefore, the slope of the tangent line to the graph of y = x^3/x - x at the point (1, -2/3) is 0, not undefined.

It's possible that you made a mistake during the derivative calculation or the substitution of the x-coordinate. Double-check your steps, and if you're still having trouble, feel free to ask for further assistance.