Differentiating polynomial absolute value function

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The solution is,

My solution is,

Where I solved these equations to find where the function is $f(x) > 0$ and $f(x) < 0$.

Using $x^2(1 - x) \geq 0$ and $x^2(1 - x) < 0$

First equation:

$x^2 \geq 0 \implies x \geq 0$ and $1 \geq x$

Second equation:

$x^2(1 - x) < 0$

$x^2 < 0 \implies x < 0$ and $1 < x$

However, I'm confused why they don't include the zero csae as I have done.

Does someone please know why this is the case?

Thanks!

 

[fresh_42]

I would prefer to say that two limits at $x=1$ exist, but are different from the right and from the left. That's a bit more precise than saying it does not exist. Of course, you are right. The derivative at $x=1$ does not exist.

I cannot see where "they" excluded $x=0$. You can use the product rule for $y=x^2\cdot |1-x|$ at $x=0.$ And for every value except $x=1.$

 

[SammyS]

. . .

Where I solved these equations to find where the function is $f(x) > 0$ and $f(x) < 0$.

Using $x^2(1 - x) \geq 0$ and $x^2(1 - x) < 0$

First equation:

$x^2 \geq 0 \implies x \geq 0$ and $1 \geq x$

Second equation:

$x^2(1 - x) < 0$

$x^2 < 0 \implies x < 0$ and $1 < x$

However, I'm confused why they don't include the zero case as I have done.

Does someone please know why this is the case?

Thanks!

(Emphasis added by me.)

Those are not equations. They are inequalities and must be treated as such. By the way; $x < 0$ and $1 < x$ is true for no value of $x$.

Yes, if you have the equation, $\displaystyle \ x^2(1 - x) = 0\,,\$ then you can say that $x^2=0$ or $x=1\ .$

However, these are inequalities . Let's look at the second inequality first.

$\displaystyle \quad \quad \ x^2(1 - x) \lt 0\,\$

You have the product of $x^2$ and $(1-x)$ . Your inequality states that this product must be negative.
Hopefully you know that $x^2\gt 0$ for all values of $x$. so the only way for the product to be negative is for $(1-x)$ to be negative (and for $x^2\ne 0$). Of course, for $(1-x)$ to be negative, we must have, $x\gt 1$ .

If $\displaystyle \ x^2(1 - x) \$ is not negative for some value of $x$, then it must be positive or zero for that value.

 

[Gavran]

Homework Statement: Please see below
Relevant Equations: Please see below

Second equation:

x2(1−x)<0

x2<0⟹x<0 and 1<x

However, I'm confused why they don't include the zero csae as I have done.

There is not a zero case because $x^2 \ge 0$ holds for any $x \in \mathbb R$.