How do you use the second derivative to determine concave up / down forf(x)=3x3−18x2+6x−29?

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despescarwh9 · Accepted Answer

Step 1At the out set, it is a cubic function. It has two turning points.When the function is minimum, the curve is concave upwards.When the function is maximum the curve is concave downwards.Find the first derivative.Set it equal to zero.It is a quadratic equation. It has two x values.Find the second derivative.Substitute the already calculated values of x to decide whether the function has a minimum or maximum.At x=3.82 The second derivative is positive. The function has a minimum. At this point the curve is concave upwards.At x=0.17. The second derivative is negative. The function has a maximum. At this point the curve is concave downwards.y=3x3−18x2+6x−29dydx=9x2−36x+6dydx=0⇒9x2−36x+6=0x=−b±b2−4ac2a=−(−36)±(−36)2−(4×9×6)2×9x=36±1296−21618=36±32.8618x=36+32.8618=68.8618=3.82x=36−32.8618=3.1418=0.17d2ydx2=18x−36At x=3.82;d2ydx2=18(3.82)−36=68.76−36=32.76&amp;gt;0At x=3.82 the function has a minimum so, the curve is concave upwardsAt x=0.17;d2ydx2=18(0.17)−36=3.06−36=−32.94&amp;lt;0At x=0.17 the function has a maximum so, the curve is concave downwards.

bobthemightyafm · Accepted Answer

Step 1f(x)=3x3−18x2+6x−29f′(x)=9x2−36x+6f″(x)=18x−36The graph of f is concave up on intervals on which f″ is positive and concave down ehere it is negative. So ingestogate the sign on f″(x)=18x−36Because this f″ is never undefined, the only chance this f″ has to change sign is when it is zero.18x−36=0 at x=2This cuts the number line into two pieces:on (−∞, 2) we have f″(x)&amp;lt;0 choose a test number if you likeon (2, ∞) we have f″(x)&amp;gt;0 choose a test number if you likeSo the graph of f is concave down on (−∞, 2) and concave up on (2, ∞)Because f(2)=−65, the point (2, −65) is the inflection point.

How do you use the second derivative to determine concave

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