• In brief, if
• f″(a) < 0: f(x) is concave at x = a
• f″(a) > 0: f(x) is convex at x = a
• If f″(x) < 0 for all x in the domain, f(x) is strictly concave.
• If f″(x) > 0 for all x in the domain, f(x) is strictly convex.
• To distinguish mathematically between a relative maximum and minimum, the second-derivative test is used. Assuming f′(a) = 0,
• 1. If f″(a) > 0, indicating that the function is convex at x =a then the function must be at a relative minimum at x = a.
• 2. If f″(a) < 0, denoting that the function is concave at x = a then the function must be at a relative maximum at x = a.
• 3. If f″(a) = 0, the test is inconclusive.

• Example 3
• Optimize f(x) = 3x³ − 54x² + 288x – 22.
• Soln.
• (a) Find the critical points, i.e f′(x) = 9 x² − 108x + 288 = 0
• Factorizing, we have 9(x − 4)(x − 8) = 0. Implies x = 4 x = 8 as the critical points.
• (b) Test for concavity by taking the second derivative, evaluating it at the critical points, and checking the signs to distinguish between a relative maximum and a relative minimum.
• f″(x) = 18x − 108
• f″(4) = 18(4) − 108 = −36 < 0 concave, relative maximum
• f″(8) = 18(8) − 108 = 36 > 0 convex, relative minimum
• The function is maximized at x = 4 and minimized at x = 8.

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• Trial
• Q1. Find the relative extrema for the following functions by
• (i) finding the critical value(s) and (ii) Determining whether at the critical value(s) the function is at a relative maximum or minimum.
• (a) f(x) = −9x² + 126x – 45
• (b) f(x) = 2x³ − 18x² + 48x – 29
• (c) f(x) = x⁴ + 8x³ − 80x² + 195
• (d) f(x) = 2x⁴ − 8x³ − 40x² + 79
•  
• Q2. (a) Find the critical values, (b) test for concavity to determine relative maxima or minima, (c) check for inflection points, (d) evaluate the function at the critical values and inflection points, and (e) graph the function, given f(x) = x³ − 18x² + 81x – 58
•  
• Q3. Redo Q2 given f(x) = −2x³ + 12x² + 72x − 70.

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