How to find the stationary point of y= x^2-108x^(1/2)+16 and determine the nature of the stationary point?

a). Find stationary point: Stationary point is the point at which the gradient equals zero. So first we must find the gradient and the set it to zero and solve: dy/dx= 2x-54x^(-1/2); now we set this to zero: 2x-54x^(-1/2)=0; 2x=54x^(-1/2) multiply both sides by x^(1/2): 2x^(3/2)=54 so x^(3/2)=27 so x^(1/2) = 3 so x=9. To find y co-ordinate plug x=9 into equation: y=81-324+16= -227 so stationary point : (9,-227) b). To determine nature of secondary point we must find the sign of the second derivative at x=9. First find second derivative: d^2y/dx^2=2+27x^(-3/2), when x=9, d^2y/dx^2=3 therefore as d^2y/dx^2>0 stationary point is a minimum.

DS
Answered by Dylan S. Maths tutor

4781 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the intersection points between the graphs y=2x+5 and y=x^2-9.


find the coordinates of the turning points of the curve y = 2x^4-4x^3+3, and determine the nature of these points


Find the derivative of the function y=3x^2e^(2x)sin(x).


How would I differentiate a function such as f(x)=x^3(e^(2x))?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences