Finding function's roots, derivatives, maxima and minima, and similar are common exercises in mathematics. For simple functions, it's easy to do it on paper. For more complex ones, not as much.

However, you'd want to check your solution in any case. A common way to do that is to use a website with the tools for functional analysis (like Wolfram Alpha), or even better, a more complete software (like Mathematica).

Such software is usually proprietary, so to use its full potential, you have to pay. Ultimately, students who are financially limited cannot use it. Another problem is that you cannot use the software for higher-level programming, which limits your future work.

An alternative way is to use a programming language suited for the job. One of the modern options is Julia, with common math functions available without imports (Math.sqrt needed no more). Here I'll present a way to do it in Julia's REPL (read-eval-print loop), which gives you the answers interactively, rather than executing a file with your code.

Plotting

Before going further with your analysis, it's useful to take a look at the function's graph. A common way to do that in Julia is with the Plots package (]add Plots to add it in Julia REPL).

We'll try to plot this polynomial:

It is as simple as:

julia> using Plots

julia> f(x) = x^3 - 6x^2 + 11x - 6
f (generic function with 1 method)
julia> plot(f)

Finding roots

The graph is there, but it isn't really precise, as you can't tell where the roots are exactly. Looking at it, you may notice they are somewhere in $[0, 4]$. To plot on range $[a, b]$ only, use plot(f, a, b). It's also useful to put the x-axis, $y = 0$ (named zero in Julia), on the same graph.

julia> plot(f, 0, 4)
julia> plot!(zero, 0, 4) # don't forget the range for zero function

There are the three roots that the polynomial has. They obviously are 1, 2 and 3, but let's check it.

Very unexpectedly, there's Roots package, too (]add Roots). It requires a range to look for the solutions, but we already have that ($[0, 4]$). It turns out our solutions were right:

julia> using Roots

julia> find_zeros(f, 0, 4)
3-element Array{Float64,1}:
 0.9999999999999999
 2.0
 3.0000000000000004

Derivatives

The third and the last package for today is Calculus (don't forget to add it). It has a simple interface for differentiating a function.

We suspect that derivative is close to zero around 1.4 (local extremum) and that second derivative is close to zero around 2.0 (inflection point).

julia> using Calculus

julia> derivative(f, 1.4)
0.08000000003876237
julia> second_derivative(f)(2.0)
-1.9021615881580974e-6

(second_derivative has to be used this way, for technical reasons.)

Let's plot the derivative, to combine all the ideas from this post:

julia> plot(derivative(f), 0, 4)

julia> plot!(zero, 0, 4)

Symbolic differentiation

Perhaps you want not only to check your derivative numerically, but also symbolically. For example, you may want to check is $\frac{d}{dx} (e^x + x^2) = e^x + 2x$. That involves a bit more work.

First, it is more convenient to define your function as a symbolic expression. You do that by adding :() around the expression:

julia> f = :(e^x + x^2)

Now using Calculus, you can get a symbolic expression for the derivative:

julia> (differentiate(f))
:(e ^ x * ((0x) / e + 1 * log(e)) + 2 * 1 * x ^ (2 - 1))

Neat, isn't it? (No, it isn't.) You can make it more neat using simplify and deparse, also provided by Calculus:

julia> simplify(ans)
:(e ^ x * log(e) + 2x)
julia> deparse(ans)
"e ^ x * log(e) + 2 * x"

ans represents the answer of the previous input, just like on your pocket calculator. It still isn't perfect, as there is an unnecessary factor log(e) (equal to 1).

Further reading

Documentation sites for the packages used in this post are very useful. There is also Calculus with Julia, a nice explanation for doing anything mathematical from $2 + 2$ up to triple integrals.