f′(x) H110054x30 H110054x30 H11005xSo, f′(x) H110050 when xH110050.Substitute xH110050 into f″(x) H1100512x2.f″(0)H11005 12(0)2f″(0)H11005 0Since f″(0) H110050, it appears that this is a point of inflection. However, the second derivative, f″(x)H11005 12x2, is always positive, so it does not change sign, and there is no change in concavity. This function is always concave up, because f″(x) is always greater than or equal to zero. H11002121460yxfx( )4xH11005(0, 0)

172MHR • Calculus and Vectors • Chapter 3Example 3 Interpret the Derivatives to Sketch a FunctionSketch a graph of a function that satisfies each set of conditions.a)f″(x) H11005H110022 for all x, f′(-3) H110050, f (H110023) H110059b)f″(x) H110210 when xH11021H110021, f″(x) H110220 when xH11022H110021, f′(H110023) H110050, f′(1) H110050Solutiona)f″(x) H11005H110022 for all x, so the function is concave down.f′(H110023) H110050, so there is a local maximum at xH11005H110023.The function passes through the point (H110023, 9).H110022H110022H110024H11002624680yx(H110023, 9)b)f″(x)H11021 0 when xH11021H110021, so the function is concave down to the left of xH11005H110021.f″(x)H110220 when xH11022H110021, so the function is concave up to the right of xH11005H110021.f′(H110023) H110050, so there is a local maximum at xH11005H110023. f′(1) H110050, so there is a local minimum at xH110051.H110022H110024H110022H110024H1100264220xyNote that this is only one of the possible graphs that satisfy the given conditions. If this graph were translated up by kunits, k H33528g92, the new graph would also satisfy the conditions since no specific points were given.

3.3 Concavity and the Second Derivative Test • MHR173<<>>KEY CONCEPTSThe second derivative is the derivative of the first derivative. It is the rate of change of the slope of the tangent.Intervals of concavity can be found by using the second derivative test or by examining the graph off″(x).• A function is concave up on an interval if the second derivative is positive on that interval. If f′(a) H110050 and f″(a) H110220, there is a local minimum at (a, f (a)).• A function is concave down on an interval if the second derivative is negative on that interval. If f′(a) H110050 and f″(a) H110210, there is a local maximum at (a, f (a)).• If f″(a) H110050 and f ″(x) changes sign at xH11005a, there is a point of inflection at (a, f (a)).Communicate Your UnderstandingC1Describe what concavity means in terms of the location of the tangent relative to the function.C2If a graph is concave up on an interval, what happens to the slope of the tangent as you move from left to right?C3When there is a local maximum or minimum on a function, the first derivative equals zero and changes sign on each side of the zero. Make a similar statement about the second derivative. Use a diagram to explain.C4Describe how to use the second derivative test to classify critical points.

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