Menu

Curve sketching

Let’s sketch the function

>> f(x_) := 4 x / (x ^ 2 + 3 x + 5)

The derivatives are

>> {f'(x), f''(x), f'''(x)} // Together
{(20-4*x^2)/(25+30*x+19*x^2+6*x^3+x^4),(-120-120*x+8*x^3)/(125+225*x+210*x^2+117*x^
3+42*x^4+9*x^5+x^6),(480+1440*x+720*x^2-24*x^4)/(625+1500*x+1850*x^2+1440*x^3+
771*x^4+288*x^5+74*x^6+12*x^7+x^8)}

To get the extreme values of f, compute the zeroes of the first derivatives:

>> extremes = Solve(f'(x) == 0, x)
{{x->-Sqrt(5)},{x->Sqrt(5)}}

And test the second derivative:

>> f''(x) /. extremes // N
{1.65086,-0.064079}

Thus, there is a local maximum at x = Sqrt(5) and a local minimum at x = -Sqrt(5). Compute the inflection points numerically, choping imaginary parts close to 0:

>> inflections = Solve(f''(x) == 0, x) // N // Chop
{{x->4.29983},{x->-1.0852},{x->-3.21463}}

Insert into the third derivative:

>> f'''(x) /. inflections
{0.00671894,-3.67683,0.694905}

Being different from 0, all three points are actual inflection points. f is not defined where its denominator is 0:

>> Solve(Denominator(f(x)) == 0, x)
{{x->1/2*(-3-I*Sqrt(11))},{x->1/2*(-3+I*Sqrt(11))}}

These are non-real numbers, consequently f is defined on all real numbers. The behaviour of f at the boundaries of its definition:

>> Limit(f(x), x -> Infinity)
0 

>> Limit(f(x), x -> -Infinity)
0

Finally, let’s plot f:

>> Plot(f(x), {x, -8, 6})
Feedback
Tell us anything that can be improved