The maximum slope of the curve y = 1/2x⁴ - 5x³ + 18x² - 19x occurs at the point :

Solution

Maths sir · 4 steps

Here the slope is the function to maximise, so differentiate it once more and test that.

1. Slope as a function. The slope at any point is m(x) = y' = 2x³ - 15x² + 36x - 19. We want the x where m is largest.

   dm/dx = 6x² - 30x + 36 = 6(x-2)(x-3)

2. Second derivative test. d²m/dx² = 12x - 30. At x = 2 it is -6 < 0, so the slope has a maximum there. At x = 3 it is +6 > 0, so the slope has a minimum there.
3. Find the point. y(2) = 1/2(16) - 5(8) + 18(4) - 19(2) = 8 - 40 + 72 - 38 = 2. The maximum slope there is m(2) = 9.
4. So. The maximum slope occurs at the point (2, 2), option (d).

The idea to keep. When a question asks for the maximum slope, treat y' as the new function: set y'' = 0 and check the sign of y'''. The point you report is still (x, y) on the curve.

Answer from the NTA final answer key. Solution by PrepWiser.