The first and second derivative test and curve sketching?

Question

In the first derivative test; If we only have one critical number, do we have an inflexion point or we don't?

Answer 1

Either one is possible. Consider y=x² (one critical value, which is not an inflection point) and y=x³ (one critical value, which IS an inflection point).

Answer 2

Not so fast,
Y = -x^4
The second derivative is zero and this is not an inflexion point.

When dy/dx = 0, the slope of the tangent to the curve is zero and thus horizontal. The curve is said to have a stationary point at a point where dy/dx = = 0.

From how y changes around the critical point we can know if it is a maximum or a minimum.

The second derivative when it is zero at certain x(i), may indicate that x(i) is an inflexion point, but not necessarely. See for example the graph of -x^4. The second derivative is zero at x=0 and zero is a maximum but not an inflexion point.

Answer 3

The first derivative only shows whether the curve is increasing(>0),decreasing(<0) or is an extremum(=0).

The second derivative must be used to derive additional information.

If the first derivative is 0, the second derivative tells whether it is an inflexion point (=0), or a maximum (< 0) or a minimum (>0).

Answer 4

The first derivative, when set = to 0 finds max, min and inflection points. To determine which we have, take the second derivative.
If it yields a positive result when evaluated at the critical point, we have a min. If it yields a negative result when evaluated at the critical point, we have a max. And if it yields 0 or an infinite value when evaluated at the critical point, that point is an inflection point.