# Power Up

Show without recourse to any calculating aid that 7^{1/2} + 7^{1/3}
+ 7^{1/4} < 7 and 4^{1/2} + 4^{1/3} + 4^{1/4} > 4 . Sketch
the graph of f(x) = x^{1/2} + x^{1/3} + x^{1/4} -x

Show without recourse to any calculating aid that:

$$7^{1/2} + 7^{1/3} + 7^{1/4} < 7$$

and

$$4^{1/2} + 4^{1/3} + 4^{1/4} > 4$$

Sketch the graph of

$$f(x) = x^{1/2} + x^{1/3} + x^{1/4} -x$$

As a hint, try comparing the '$7$'s' inequality to a similar one for $8$. For the '$4$'s' inequality use the fact that any root of $4$ is greater than $1$.

To sketch the graph, find the derivative for $x=0$ and then consider where the derivative is positive, where it is negative and if it tends to a limit as $x$ increases.

Graeme showed the inequalities for us:

$7 = 9^{1/2} + 8^{1/3} + 16^{1/4} > 7^{1/2} + 7^{1/3} + 7^{1/4}$

$4 = 4^{1/2} + 1^{1/3} + 1^{1/4} < 4^{1/2} + 4^{1/3} + 4^{1/4}$

While I was at it, I came up with these:

$6 = 6.25^{1/2} + 6.859^{1/3} + 6.5536^{1/4} > 6^{1/2} + 6^{1/3} + 6^{1/4}$

Although that looks hard, it can be done without a calculator by partitioning $6$ into $2.5+1.9+1.6$, and finding appropriate powers of each number. The last one is easy for computer geeks like me who have memorized many small powers of $2$.

$5 < 4^{1/2} + 4.096^{1/3} + 4^{1/4} < 5^{1/2} + 5^{1/3} + 5^{1/4}$

This, too, is pretty easy without a calculator - $4.096^{1/3}$ is $1.6$, and the square root of $2$ is more than $1.4$, so the first sum is more than $5$, and clearly less than the second sum.

Thanks for the extensions, Graeme.

These inequalities show that the graph is going to intersect the x-axis somewhere between 4 and 7, which it does:

Image