## University Calculus: Early Transcendentals (3rd Edition)

The equation has 1 solution: $(1.56)$
$$x^x=2$$ $$x^x-2=0$$ To prove the equation has a solution using the Immediate Value Theorem, we need to find 2 values $x_1$ and $x_2$ and show that there is a change of sign in the function $f(x)=x^x-2$ as $x$ goes from $x_1$ to $x_2$. - Take $x_1=1$: $f(1)=1^1-2=1-2=-1\lt0$ - Take $x_2=2$: $f(2)=2^2-2=4-2=2\gt0$ So there is a change of sign in $f(x)$ as $x$ goes from $1$ to $2$. - On $[1,2]$: $\lim_{x\to c}f(x)=\lim_{x\to c}(x^x-2)=c^c-2=f(c)$ So $f(x)$ is continuous on $[1,2]$. Therefore, according to the Intermediate Value Theorem, there must be a value $x=c\in[1,2]$ such that $f(c)=0$. In other words, the equation $x^x-2=0$ or $x^x=2$ has at least one solution in the interval $[1,2]$. The graph of the function $f(x)=x^x-2$ is enclosed below. Looking at the graph, it turns out that the curve $f(x)$ crosses the line $y=0$ at only $1$ point whose $x\in[1,2]$. The $x$-coordinate of that point is the solution of the equation. So the equation has 1 solution: $(1.56)$