Show that $f(x)=x^{5/3}-kx^{4/3}+k^2x$ is increasing for $k\neq0$

Question

So to show the function is increasing/decreasing we differentiate and show it is more than zero/less than zero: We have

$$f(x)=x^{5/3}-kx^{4/3}+k^2x$$

Hence,

$$f'(x)=\frac{5}{3}x^{2/3}-\frac{4k}{3}x^{1/3}+k^2$$

But how do I show

$$\frac{5}{3}x^{\frac{2}{3}}-\frac{4k}{3}x^{\frac{1}{3}}+k^2>0$$

Answer 1

Change variables to $y = x^{1/3}$ to get the quadratic equation $$5y^2 - 4ky + 3k^2>0$$ with discriminant $$ b^2-4ac = 16k^2-60k^2 $$ Can you finish this?

Answer 2

Let

$$y = x^{\frac{1}{3}}$$

So,

$$\frac{5}{3}y^{2} - \frac{4k}{3}y + k^{2} > 0 \Longleftrightarrow 5y^{2} - 4ky + 3k^{2} > 0$$

and

$$(-4k)^{2} - 4.5.3k^{2} = 16k^{2} - 60k^{2} < 0.$$

Therefore, $\frac{5}{3}y^{2} - \frac{4k}{3}y + k^{2} > 0$. Complete the solution.

Answer 3

Notice that you now have a new second order polynomial equation with $k$ as a variable instead of x.

You can compute the discriminant $\Delta$: $$\Delta = (\frac{4x^{\frac{1}{3}}}{3})^2 - 4\frac{5x^\frac{2}{3}}{3} $$ $$\Delta = \frac{16x^{\frac{2}{3}}}{9} - \frac{60x^\frac{2}{3}}{9} = -\frac{44x^\frac{2}{3}}{9} < 0$$ Therefore $f'$ has no root and has the same sign as $k^2$