You may also like Real(ly) Numbers

If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have? How Many Solutions?

Find all the solutions to the this equation. After Thought

Which is larger cos(sin x) or sin(cos x) ? Does this depend on x ?

Exponential Trend

Age 16 to 18 Challenge Level: Well done Curt from Reigate College for this solution.

If $G(x)=e^{f(x)}$, by the chain rule:

$$G'(x)=e^{f(x)} \times f'(x).$$

As $e$ is never zero, the gradient function of $G(x)$ is zero when the gradient function of $f(x)$ is zero and it always has the same sign as $f'(x)$ so $G(x)$ and $f(x)$ have the same turning points.

To answer the question of turning points of $G(x)=x^{1\over x}= e^{1/x\log x}$ let's first find the derivative of ${1\over x}\log x$.

$${d\over dx }f(x)={d\over dx }({1\over x}\log x) = {{1-\log x}\over x^2}.$$

The derivative is positive for $\log x < 1$, that is $x< e$, it is is zero when $\log x = 1$, $x = e$ and it is negative for $x> e$. Hence the function is increasing for $x< e$ and decreasing for $x> e$.

$${d^2\over dx^2 }({1\over x}\log x)= {{-x-2x(1-\log x)}\over x^4 }= {{2\log x -3}\over x^3}.$$

At $x=e$ the second derivative of $f(x)$ is $-1/e^3$ which is negative so $f(x)$ has a maximum and so $G(x)$ has a maximum. Therefore $(e, e^{1/e})$ is a maximum point of $G(x)=x^{1\over x}$.

To prove $\lim x^{1\over x} \rightarrow 1$ as $x\rightarrow \infty$ we may use the similar result for the discrete case $n^{1\over n}$. As the graph of $f(x)=x^{1\over x}$ is decreasing as $x\to \infty$ for each value of $x$ there is a value of $n$ for which $n^{1/n}< x^{1/x}$ and vice versa so both tend to the same limit which is $1$.

What if $x\rightarrow 0$? Well, as suggested in the question, letting $1/x=t$ in $x^{1\over x}$, one can note that as $x\rightarrow 0$, $t\rightarrow \infty$. Writing $t=1/x$, $$x^{1/x} = \left ({1\over t}\right )^t = {1\over t^t}.$$ As $x\to 0$ we have $t\to \infty$ and $t^t \to \infty$ so $$x^{1/x} = {1\over t^t} \to 0$$.

Hence the graph of $y=x^{1/x}$ is always above the line $y=0$ although the function is undefined at $x=0$.