Why can f(x) = x^2 never be negative but f(x) = x^2.2 can?

[molten radish]

Both can't give negative values when $x \in \mathbb{R}$. This is because of the fact that $$x^2 = x\cdot x$$ and this is always positive. \ For $x^{2.2}$ we have to rewrite it for it to become clear that it can't give negative values. So we know that $$2.2 = 2\cdot 1.1$$ therefore $$x^{2.2} = x^{2\cdot 1.1} = (x^{1.1})^2$$ So we get something of the form $a^2$ which, as mentioned before, can't be negative.

[zinc topaz]

zlik

[molten radish]

Does this help?

[foggy atlas]

It would, but it's flawed

First, rewrite the exponent as a fraction.
$$x^{2.2}=x^{\frac{22}{10}}$$
Then take the fraction and write it as a combination of root and exponent.
$$x^{\frac{22}{10}}=\sqrt[10]{x^{22}}$$
Simplify.
$$\sqrt[10]{x^{22}}=\sqrt[2(5)]{x^{2(11)}}=\sqrt[5]{x^{11}}$$
Now, check all three possibilities for $x$. I'll write $n$ for negative numbers and $p$ for positive numbers.$\$
$x<0$:
$$\sqrt[5]{n^{11}}=\sqrt[5]{n}=n$$
$x=0$:
$$\sqrt[5]{0^{11}}=\sqrt[5]{0}=0$$
$x>0$:
$$\sqrt[5]{p^{11}}=\sqrt[5]{p}=p$$

[zinc topaz]

animeonfire

[foggy atlas]

While your argument of squaring $x^{1.1}$ seems valid at first, you have overlooked the possibility of $x^{1.1}$, for $x\in\mathbb{R}, x<0$ to be complex.

[zinc topaz]

animeonfire

[foggy atlas]

I hope this helps both of you.

[molten radish]

Can you give me an example of such x?

[foggy atlas]

It goes for any negative $x\in\mathbb{R}$

[zinc topaz]

animeonfire

[foggy atlas]

Let's do x=-1 for demonstration purposes

$$(-1)^{1.1}=(-1)^{\frac{11}{10}}=\sqrt[10]{(-1)^{11}}$$
$$=\sqrt[2]{\sqrt[5]{-1^{11}}}=\sqrt[2]{-1}=i$$

[zinc topaz]

animeonfire

[foggy atlas]

Therefore

$((-1)^{1.1})^{2}=i^{2}=-1$

[zinc topaz]

animeonfire

[molten radish]

Oooh yes, makes sense. I don't know how I looked over that

Thanks!

[foggy atlas]

.solved