# Question #60f48

Jun 4, 2016

The easiest way of doing this is using a calculus concept called the intermediate value theorem. The intermediate value theorem basically states that given a function $f \left(x\right)$ which is continuous on an interval $\left[a , b\right]$, $f$ takes on every value between $f \left(a\right)$ and $f \left(b\right)$ somewhere in that interval.

While a function being continuous has a more rigorous and nuanced definition, a non-calculus student may think of it as being a function which can be graphed without lifting one's pen off of the paper. This suffices for most simple functions, and also gives some idea as to why the intermediate value theorem works. If you never take your pen off the paper, then to get from $f \left(a\right)$ to $f \left(b\right)$, you need to pass over each point between them.

As it happens, a polynomial function will be continuous on any interval, and we can use that to show that the given equation has a solution. Consider, for example, the interval $\left[- 2 , 1\right]$.

Letting $f \left(x\right) = {x}^{9} + 3 x + 15$, we have

$f \left(- 2\right) = {\left(- 2\right)}^{9} + \left(- 2\right) 3 + 15 = - 512 - 6 + 15 = - 503$

and

$f \left(1\right) = {1}^{9} + \left(1\right) 3 + 15 = 1 + 3 + 15 = 19$

As $f \left(- 2\right) = - 503$ and $f \left(1\right) = 19$, the intermediate value theorem lets us say that for any value $y$ between $- 503$ and $19$, there exists some $c \in \left[- 2 , 1\right]$ such that $f \left(c\right) = y$.

In particular, as $- 503 < 0 < 19$, we know that there exists some $c \in \left[- 2 , 1\right]$ such that $f \left(c\right) = {c}^{9} + 3 c + 15 = 0$. Then $c$ is a solution to the given equation, and so at least one solution exists.

If we look at the graph, the argument boils down to noting that as the function crosses the line $y = 0$ somewhere between $- 2$ and $1$, the point where it crosses must be a solution.

graph{x^9+3x+15 [-51.9, 52.1, -22.96, 29]}