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# The Sign of the Second Derivative

## Maxima and Minima

We have seen previously that the sign of the derivative provides us with information about where a function (and its graph) has stationary points. We now look at local maxima and minima again.

Look at the applet below. Use the slider bar to change the value of x. Two examples are given of higher order polynomials and you can enter other coefficients including 0, to totally convince yourself that:

• At all local maxima, f '(x) = 0 and f "(x) < 0.
• At all local minima, f '(x) = 0 and f "(x) > 0.
• At all points of inflection, f "(x) = 0.

This leads us to the preferred method for determining the local maxima or minima for some function.

1. Differentiate the function, f(x), to obtain f '(x) and f "(x).
2. Solve the equation f '(x) = 0 for x to get the values of x at potential minima or maxima.
3. For each x value:
1. Determine the value of f "(x) at the x value.
2. Decide if you have a minimum or a maximum.
1. At all local maxima, f "(x) < 0.
2. At all local minima, f "(x) > 0.
3. If f "(x) = 0, do the first derivative test.
3. Calculate the value of the function at the x value.

### Example

Consider the function f(x) = 5 - sin x.

Differentiating gives f '(x) = -cos x and f "(x) = sin x.

Solving f '(x) = 0 = -cos x leads to x = - or x = or generally x = + where n is an integer.

f "(x) = sin x, so f "(-) = -1 and f(-) = 6 so (-,6) is a local maximum, recurring at every (2,4).

f "() = 1 and f() = 4 so (,4) is a local minimum, recurring at every (2 + ,4).

Consider f (x) = x4. f '(x) = 4x3, so f '(x) = 0 when x = 0. f "(x) = 12x2, so f "(0) = 0. If x) < 0 then f '(x) is negative and if x) > 0 then f '(x) is positive. f '(x) is increasing. This is a local minimum. f (0) = 04 = 0 giving (0,0) is the loacl minimum point.

### Exercise

 Find the local maximum or minimum points on the graph of f(x) = x3 + x2 + x + Working: Local Maximum point(s) Local Minimum point(s)