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
Age
16 to 18
Challenge level
Problem
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
$$f(x) = x^{1/2} + x^{1/3} + x^{1/4} -x$$
Getting Started
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.
Student Solutions
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
