Given the function f(x)=-3x^4 + 8x^3 + 36x^2?

Question

..i). find the stationary points of the function

..ii). find the range of values of x for which the function is increasing

..iii). state which stationary points are maxima and minima

Answer 1

f(x) = -3x⁴+8x³+36x²

f'(x) = -12x³+24x²+72x

f'(x) = 0 -> Find stationary points

-12x³+24x²+72x = 0

x³-2x²-6x = 0

x(x²-2x-6) = 0

x=0

x²-2x-6=0

x=0

x = 2+/-rad(28)/2 = 2+/-2rad7 /2 = 1+/-rad7

x₁ = 0

x₂ = 1+√7

x₃ = 1-√7

For x=0 f(x) = -3*0+8*0+36*0 = 0

O(0;0)

For x=1+√7 f(x) = 56√7+188

A(1+√7; 56√7+188)

For x=1-√7 f(x) = 188-56√7

B(1-√7 ; 188-56√7)

O(0;0)

A(1+√7 ; 56√7 + 188)

B(1-√7 ; 188-56√7)

For knowing if we have maxima or minima points you must find f''(x)

f''(x) = -36x²+48x+72

Substitute

x = 0

f''(0) = 72

For x=0

f'(x) = 0 , f''(x) > 0

In O(0;0) we have a minima point

Substitute

x = 1+√7

f''(1+√7) = -24√7-168

For x=1+√7

f'(x) = 0 , f''(x) < 0

In A(1+√7 ; 56√7+188) we have a maxima point

Substitute

x = 1-√7

f''(1-√7) = 24√7-168

For x=1-√7

f'(x) = 0 , f''(x) < 0

In B(1-√7 ; 188-56√7) we have a maxima point

Then

O(0;0) -> minima point

A(1+√7; 56√7+188) maxima point

B(1-√7 ; 188-56√7) maxima point

f'(x) = -12x³+24x²+72x

f'(x) > 0

-12x³+24x²+72x

-12x(x²-2x-6) > 0

-12x>0 , x²-2x-6 > 0

x<0 , x<1-√7 v x>1+√7

x<1-√7

-12x<0 , x²-2x-6 <0

x>0 , 1-√7 < x < 1+√7

0<x<1+√7

Then for

x<1-√7 v 0<x<1+√7

The function is increasing

Answers

i) , ii)

O(0;0) -> minima point

A(1+√7; 56√7+188)-> maxima point

B(1-√7 ; 188-56√7)-> maxima point

iii)

for

x<1-√7 v 0<x<1+√7

The function is increasing