Question

For the function f(x)=4x^3-x^4

1. Find all critical numbers and corresponding critical points.

2. find relative (or called local) extremum and corresponding extreme points.

3.Find the points of inflection, intervals of concavity and sketch a graph.

Answer 1

first differentiate with respect to x
f'(x) = 12x^2 - 4x^3
equate this to 0 and solve for x

Answer 2

i dont even know what they are asking for or how to find it

Answer 3

the derivative 12x^2 - 4x^3 gives the slope of the tangent at the poin(x,f(x) on the curve of the function The critical pointa are where the curve turns to give a maximum minimum or poit of inflection At these points the derivative f'(x) = 0.

so we solve 12x^2 - 4x^3 = 0
4x^2(3 - x) = 0
so x = 0 or 3 at critical points

Answer 4

i have to leave for a while sorry - be back later

Answer 5

thanks for help :)

Answer 6

oh and patrick jmt i have added on my utube page, he was awesome for my trig class

Answer 7

now to find the nature of the critical points first find the second derivative
y" = 24x - 12x^2
when x = 3, y" = 72 - 108 - negative so this is a maximum

when x = 0 y" = 0 so we need to use y'
we need to take values of x just below 0 and just above 0.
let x = -0.1: y' = 12(-0.1)^2 - 4 (-0.1)^3 = 0.12 + 4* 0.001 which is positive
let x = 0.1 y' = 0.12 - 0.004 which is also postive
this is a point of inflection

coordinats of local maximum are (3,(4(3)^3 - 3^4) = (3, 27)
and coodirnates of point of inflection is (0,0)

Answer 8

ok - i'll try and draw a rough sketch !!