Concavity We ve seen how knowing where a unction is increasing and decreasing gives a us a good sense o the shape o its graph We can reine that sense o shape by determining which way the unction bends This bending is called the concavity o the unction I am going to use a dierent deinition o concavity than the text But the deinition in the text will be a consequence (or theorem) resulting rom my deinition So they end up the same Curves that bend up so that they are cupped up or hold water are called concave up while curves that bend down or are cupped down so that they spill water are called concave down Parabolas provide the basic examples Figure 321: Concave up (let) and concave down curves Note where the tangents are in relation to each curve Then note how the slope o the tangent lines is changing or each curve Notice the location o the tangents to each type o curve DEFINITION 321 I all o the tangents lie below the graph o y = (x) on an interval I, then is concave up on I I all o the tangents lie above the graph o y = (x) on an interval I, then is concave down on I Now look at what has to happen to the slope o or it to be concave up: The slope (ie, (x)) must be increasing so its derivative must be positive That is Concave up slope o is increasing (x) is increasing d dx ( (x)) > (x) > Concave down slope o is decreasing (x) is decreasing d dx ( (x)) < (x) < This gives us
math 13, day 32 concavity 2 THEOREM 321 (The Concavity Test) Let be a unction whose second derivative exists on an interval I 1 I (x) > or all points in I, then is concave up on I 2 I (x) < or all points in I, then is concave down on I Concave up Concave down Concave up Concave down Concave up Notice in the curve above that there are several places where the concavity switches The second derivative must be (or else it does not exist) at the point since the curve is not bending either way We give these points where the concavity changes a special name DEFINITION 322 A point P is called a point o inlection or i changes concavity there We can ind such points by looking or places where the second derivative, (x), changes sign EXAMPLE 321 Find the intervals o concavity and the inlection points or (x) = x 4 6x 2 + 1 Sketch a graph that includes relative extrema, critical numbers, and inlections SOLUTION Begin with the irst derivative Previously we saw (x) = 4x 3 12x = 4x(x 2 3) = 4x(x 3)(x + 3) = at x = ± 3, dec l min inc l min dec l max inc + + + + + + (3 1/2 ) 3 1/2 Now we can take the second derivative (x) = 12x 2 12 = 12(x 2 1) = at x = ±1 These are the potential inlection points; we still have to determine whether the sign o actually changes at these points (so that the concavity changes) concave up inlect conc dn inlect concave up + + ++ + + ++ To make a graph we will plot just the critical numbers and the inlections and then connect them appropriately based on the inormation rom the two derivatives ( 3) = 9 18 + 1 = 8 = ( 3) = 9 18 + 1 = 8 () = 1 (1) = 1 6 + 1 = 4 = ( 1) r max (3) 1/2 3 1/2 inl 4 inl 8
3 Note The various combinations o increasing/decreasing concave up/down are illustrated below dec, conc up inc, conc up inc, conc dn dec, conc dn The graph indicates that or smooth (dierentiable twice) unctions, that at a local min the unction must be concave up and at a local max the unction is concave down Interpreting concavity by using the second derivative leads to THEOREM 322 (Second Derivative Test) Assume that is a unction so that (c) = and exists on an open interval containing c (a) I (c) >, then has a local min at c (b) I (c) <, then has a local max at c (c) I (c) =, then the test ails there may be a local max, o, or neither at c Possibility (3) is why the First Derivative Test is more useul when classiying critical numbers In this regard, the Second Derivative Test acts mostly as a check to ensure that you have not made an error in classiication YOU TRY IT 321 From Test 3A: Do a complete graph o (x) = x 4 + 4x 3 + 1 Indicate all extrema, inlections, etc EXAMPLE 322 Do a complete graph o y = (x) = x 5 5x 4 Indicate all extrema, inlections, etc SOLUTION Critical numbers: (x) = 5x 4 2x 3 = 5x 3 (x 4) = at x =, 4 inc r max dec inc + + + + + + Now take the second derivative 4 (x) = 2x 3 6x 2 = 2x 2 (x 3) = at x =, 3 These are the potential inlection points; we still have to determine whether the sign o actually changes at these points conc dn No In conc dn Inlect concave up + + ++ 3 Aside: Apply the second derivative test to the critical numbers Notice that (4) >, thereore by the second derivative test there is a local min at x = 4 However, () =, so the second derivative test ails Nonetheless, the irst derivative test tell us that there is relative max at To make a graph we will plot just the critical numbers and the inlections and then connect them appropriately based on the inormation rom the two derivatives (a) () = (b) (4) = 124 128 = 256 (c) (3) = 243 45 = 162
math 13, day 32 concavity 4 r max 3 4 162 256 inl EXAMPLE 323 Do a complete graph o y = (x) = inlections, etc x2 Indicate all extrema, x 2 +1 SOLUTION First locate critical numbers: (x) = 2x(x2 + 1) x 2 (2x) 2x (x 2 + 1) 2 = (x 2 = at x = + 1) 2 dec l min inc + + ++ Check concavity with the second derivative (x) = 2(x2 + 1) 2 2x(2)(x 2 + 1)(2x) (x 2 + 1) 4 = 2(x2 + 1) 8x 2 (x 2 + 1) 3 = 2 6x2 (x 2 + 1) 3 = at x = ±1/ 3 ±577 concave down inlect conc up inlect concave down + + + (1/3) 1/2 (1/3) 1/2 Plot the critical numbers and the inlections and then connect them appropriately based on the inormation rom the two derivatives (a) () = (b) ( 1/ 3) = 1/3 4/3 = 1/4 (c) (1/ 3) = 1/4 inl 1/4 inl (1/3) 1/2 (1/3) 1/2 EXAMPLE 324 Do a complete graph o y = (x) = e x2 /2 Indicate all extrema, inlections, etc SOLUTION Locate critical numbers: (x) = xe x2 /2 = at x = inc l max + + ++ Check concavity with the second derivative dec (x) = e x2 /2 xe x2 /2 ( x) = (x 2 1)e x2 /2 = at x = ±1 concave up inlect conc dn inlect concave up + + ++ + + ++ Plot the critical numbers and the inlections and then connect them appropriately based on the inormation rom the two derivatives (a) () = e = 1
5 (b) ( 1) = e 1/2 67 (c) (1) = e 1/2 67 r max inl inl EXAMPLE 325 Here s a quick one with an interesting point: Do a complete graph o y = (x) = x 3 + x SOLUTION Locate critical numbers: (x) = 3x 2 + 1 = ; there are no critical numbers inc + + + + + + + + + + ++ Check concavity with the second derivative: (x) = 6x = at x = conc down inlect concave up + + ++ The only point to plot is () = The shape o the curve is determined by the irst and second derivatives inl YOU TRY IT 322 Graph each o the ollowing: (a) (x) = 1 (b) (c) Hint: The graph will look very similar to the exponential graph we x 2 +1 did above (x) = x 4 4x 3 Hint: It has critical numbers at x = and 3 and inlections at x = and 2 (x) = x cos x It has critical numbers but no extrema It has lots o inlections (d) Challenge: Graph (x) = (x 3 8) 1/3 EXAMPLE 326 Here s another quick one with a twist: Do a complete graph o y = (x) = 3x 2/3 x SOLUTION Locate critical numbers: (x) = 2x 1/3 1 = 2 1 =, so 2 x 1/3 1 = 1 x 1/3 2 so x1/3 = 2 or x = 8 Also (x) DNE at x = decrease increase r max decrease DNE + + ++ 8 x 1/3 = 1 so
math 13, day 32 concavity 6 Check concavity with the second derivative: (x) = 2/3 =, but (x) DNE at x 4/3 x = conc down DNE concave down The only points to plot is () = and (8) = 4 The shape o the curve is determined by the irst and second derivatives r max EXAMPLE 327 Below I give you inormation about the irst and second derivatives o two continuous unctions For each, sketch a unction that would have dervatives like those given Indicate on your graph which points are local extrema and which are inlections indicates that the derivative does not exist at the point (though the original unction does) You will need to make up values or the critical and inlection points consistent with the inormation supplied (a) ++ 2 2 ++ + + + 2 (b) 8 + + + + + + g ++ g 1 3 g