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Additional resources for MEI C2 Study Resources Core2 Differentiation 1-2
Example text
I) 24 x x 24 - 2x x 15 - 2x 15 x Height of box is x cm Length of box is (24 – 2x) cm Width of box is (15 – 2x) cm Volume V = x(15 − 2 x )(24 − 2 x ) = x(360 − 78 x + 4 x 2 ) = 4 x 3 − 78 x 2 + 360 x dV = 12 x 2 − 156 x + 360 dx At turning points, 12 x 2 − 156 x + 360 = 0 (ii) x 2 − 13 x + 30 = 0 ( x − 3)( x − 10) = 0 x = 3 or x = 10 © MEI, 11/08/08 6/7 MEI C2 Differentiation Section 2 Exercise solutions x = 10 is not possible since this would mean that the width would be negative. d2V = 24 x − 156 dx 2 d2V When x = 3, = 72 − 156 < 0 , so x = 1 is a maximum point.
Find the co-ordinates of any turning points and determine their nature showing clearly how your decisions were made. Sketch the curve. 4. The equation of a curve is y = (x +1)(x – 3)3. (i) Write the equation of the curve in the form y = ax4 + bx3 + cx2 + dx + e. dy (ii) Find the co-ordinates of the points where = 0. dx (iii) Classify the stationary points. (iv) Sketch the curve. 5. Find the stationary points on the curve y = x4 – 2x3 and distinguish between them, showing all of the relevant working clearly.
Iv) y = x 3 + 6 x 2 + 9 x = x( x 2 + 6 x + 9) = x( x + 3)2 The graph cuts the x-axis at x = 0 and x = -3 (repeated). The graph cuts the y-axis at y = 0. -3 (-1, -4) © MEI, 11/08/08 1/7 MEI C2 Differentiation Section 2 Exercise solutions 2. (i) y = 2 x + x 2 − 4 x 3 dy dx = 2 + 2 x − 12 x 2 dy =0 dx 2 + 2 x − 12 x 2 = 0 At turning points, 1 + x − 6x 2 = 0 6x 2 − x − 1 = 0 (3 x + 1)(2 x − 1) = 0 x = − 31 or x = 21 2 3 4 = When x = − 31 , y = 2 ( − 31 ) + ( − 31 ) − 4 ( − 31 ) = − 23 + 19 + 27 When x = 21 , y = 2 ( 21 ) + ( 21 ) − 4 ( 21 ) = 1 + 41 − 21 = 2 3 11 The turning points are ( − 31 , − 27 ) and x< − 31 -ve x dy x = − 31 0 −18 + 3 + 4 27 11 = − 27 3 4 ( 21 , 43 ) .