Question

PLEASE HELP ME!!
let g be the function given by g (x) = 3x^4 - 8x^3. At what value of x on the closed interval [ -2,2 ] does g have an absolute maximum

Answer 1

Take the derivative:
g’(x) = 12x^3 - 24x^2

Set equal to zero and solve:
0 = 12x^3 - 24x^2
0 = 12x^2 (x - 2)

x = 0 or x = 2

Plug back into original
g(0) = 3(0^4) - 8(0^3)
g(0) = 0 - 0
g(0) = 0

g(2) = 3(2^2) - 8(2^3)
g(2) = 3(4) - 8(8)
g(2) = 12 - 64
g(2) = -52

There is an absolute max at (0,0) or when x = 0

Answer 2

The value of x on the closed interval [ -2,2 ] that g has an absolute maximum is 0

Given the function $g (x) = 3x^4 - 8x^3$

The function is at a maximum at g'(x) = 0

Differentiating the function given:

$g'(x)= 12x^3-24x^2\\0= 12x^3-24x^2\\ 12x^3-24x^2=0\\12x^2(x-2)=0\\12x^2=0 \ and \ x - 2 = 0\\x = 0 \ and\ 2$

Substitute x = 0 and x = 2 into the function;

g(0) = 3(0)^3 - 8(0)^3

g(0) = 0

Since the range of the function is least at x = 0, hence the value of x on the closed interval [ -2,2 ] that g has an absolute maximum is 0

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