Graph the function. Label all intercepts, critical points, and inflection points.
f (x) = x2ex
• Find dots.
The function f (x) is only zero when x = 0.
The derivative is
which is zero when x = 0 and when x = -2, so these are our critical points. When x = 0, we have y = 0 also.
f(-2) = 4e-2
so the complete coordinates of the critical points are
(0,0) and (-2,4e-2).
For the sake of graphing,
4e-2 ≅ 0.54.
The second derivative (using the first definition of f', not the factored version) is
This is zero when (x2 + 4x + 2) is zero. We need the quadratic equation to solve this one:
We need to find the corresponding y-values.
• Find shapes.
Set up the numberline:
The function f is never negative.
The function f '(x) = xex(2 + x) is negative when exactly one of the quantities x and (2 + x) is negative, and positive otherwise. Therefore f ' is only negative when -2 < x < 0.
The function f "(x) = ex(x2 + 4x + 2) is negative when x is between the roots of the polynomial (x2 + 4x + 2). That is, when
These are the shapes of f :
• Play connect-the-dots.
We play connect-the-dots, keeping in mind that f can't go below the x-axis: