Find the stationary points of the curve y=2*x^3-15*x^2+24*x+17. Determine whether these points are maximum or minimum.

First, differentiate and put the derivative equal to zero. dy/dx=6x^2-30x+24=0. Solve this equation to get that x=4 and x=1. Substitute these values into the original equation to get the corresponding values of y. The stationary points are (1,17) and (4,-10). Calculate the second derivative to get d^2y/dx^2=12*x-30. When x=1 the second derivative is less than zero so (1,17) is a maximum point and when x=4 the second derivative is greater than zero so (4,-10) is a minimum point.

SM
Answered by Shaun M. Maths tutor

4073 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

f(x) = e^(sin2x) , 0 ≤ x ≤ pi (a). Use calculus to find the coordinates of the turning points on the graph of y = f(x)


The curve, C has equation y = 2x^2 +5x +k. The minimum value of C is -3/4. Find the value of k.


Find the set of values of x for which 3x^2+8x-3<0.


The straight line with equation y = 3x – 7 does not cross or touch the curve with equation y = 2px^2 – 6px + 4p, where p is a constant. Show that 4p^2 – 20p + 9 < 0.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning