Find the differential of f(x)=y where y=3x^2+2x+4. Hence find the coordinates of the minimum point of f(x)

The differential of f(x) is f'(x)=6x+2 because we apply this basic formula: the differential of kxn = knxn-1 where k is a constant.This is a positive quadratic, so it has a U shape that is above the x-axis (I would draw a sketch of the graph at this point). This means it has only got one point where the gradient is zero. This must be the minimum point of f(x) (I would demonstrate this on the sketch of the graph).Therefore since f'(x) tells us the gradient of f(x), the minimum point will be where f'(x)=0, so we can say 6x+2=0.Solving for this gives x=-1/3. To find the y value for the minimum point we calculate f(-1/3) by substituting x=-1/3 back into the original equation y=3x2+2x+4. This gives a value of 11/3.Hence the minimum point of f(x) is (-1/3, 11/3)

Answered by Maths tutor

3084 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the derivative of the function f:(0,oo)->R, f(x)=x^x.


Find the derivative with respect to x and the x-coordinate of the stationary point of: y=(4x^2+1)^5


Find the x value of the stationary points of the graph y = x^2e^x


How to transform graphs of functions?


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences