Which of the following functions have a derivative at x=0?

I. y= absolute value(x^3-3x^2)

II y= square root(x^2+.01)- absolute value (x-1)

III y= e^x/cosx

What is your thinking? Look to see where there are discontiuties.

I think all three of them have derivatives because when I put them into the calculator, I saw no discontinuities. but i am debating if it's II and III.

doesn't one change slope rapidly at x=0?

I think it does. For answer choice one correct?

Well, let's get to the root of the problem!

For function I, y = |x^3 - 3x^2|, let's see if it has a derivative at x = 0. The absolute value function |x| has a sharp corner at x = 0, which means its derivative does not exist at that point. Since the function I has absolute value in it, it also won't have a derivative at x = 0.

Now, for function II, y = √(x^2 + 0.01) - |x - 1|, let's evaluate its derivative at x = 0. The square root function, √x, has a derivative at x = 0, so that's good news! However, the absolute value function still has a sharp corner at x = 1, meaning its derivative doesn't exist at that point. So overall, function II won't have a derivative at x = 0 either.

Lastly, for function III, y = e^x/cosx, we need to check if it has a derivative at x = 0. Both the exponential function, e^x, and cosine function, cosx, have a derivative at x = 0. However, since we are dividing e^x by cosx, we need to make sure that cosx ≠ 0 at x = 0. Luckily, cosx ≠ 0 at x = 0, so function III has a derivative at x = 0.

So, to summarize, only function III, y = e^x/cosx, has a derivative at x = 0. The other two, functions I and II, don't have a derivative at x = 0 or should I say they "can't derive without a few x-tra twists and turns!"

To determine which of the given functions have a derivative at x=0, we need to check if the functions are continuous and differentiable at that point.

Let's go through each function:

I. y = |x^3 - 3x^2|
To determine if this function has a derivative at x=0, we need to check if it is continuous at that point. Since the function involves the absolute value, we need to consider both the left and right limits at x=0.
The left limit as x approaches 0 is: lim(x->0-) (|x^3 - 3x^2|) = 0
The right limit as x approaches 0 is: lim(x->0+) (|x^3 - 3x^2|) = 0

Since both the left and right limits are equal to 0, the function is continuous at x=0.

To determine if the function has a derivative at x=0, we also need to check if it is differentiable. Taking the derivative of the function, we get:

y' = d/dx (|x^3 - 3x^2|)
= d/dx (x^3 - 3x^2) (since the derivative of |x| is x for x ≠ 0)
= 3x^2 - 6x

Now, evaluating the derivative at x=0, we get:
y'(0) = 3(0)^2 - 6(0) = 0

Since the derivative at x=0 exists and is equal to 0, function I has a derivative at x=0.

II. y = √(x^2 + 0.01) - |x - 1|
To determine if this function has a derivative at x=0, we need to check if it is continuous at that point. Since the function has square roots and absolute value, we again need to consider the limits from the left and right at x=0.
The left limit as x approaches 0 is: lim(x->0-) (√(x^2 + 0.01) - |x - 1|) = √(0.01) - 1 = -0.01 (negative value)
The right limit as x approaches 0 is: lim(x->0+) (√(x^2 + 0.01) - |x - 1|) = √(0.01) - 1 = -0.01 (negative value)

Since the left and right limits are not equal to each other, the function is not continuous at x=0. Therefore, the function does not have a derivative at x=0.

III. y = e^x / cosx
To determine if this function has a derivative at x=0, we need to check if it is continuous at that point. In this case, we only need to evaluate the function at x=0.
y(0) = e^0 / cos(0) = 1 / 1 = 1

Since the function evaluates to a finite value at x=0, it is continuous at that point.

To determine if the function has a derivative at x=0, we also need to check if it is differentiable. Taking the derivative of the function, we get:

y' = d/dx (e^x / cosx)
= (e^x * cosx - e^x * (-sinx)) / cos^2x
= (e^x * (cosx + sinx)) / cos^2x

Now, evaluating the derivative at x=0, we get:
y'(0) = (e^0 * (cos(0) + sin(0))) / cos^2(0) = (1 * (1 + 0)) / 1 = 1

Since the derivative at x=0 exists and is equal to 1, function III has a derivative at x=0.

In summary, functions I and III have a derivative at x=0, while function II does not.