# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for f(x)= 3x^4+4x^3-12x^2+5?

Dec 10, 2016

Tthe function has a maximum at $x = 0$
The function has a minimum at $x = 1$
The function has a minimum at $x = 2$

#### Explanation:

Given -

$f \left(x\right) = 3 {x}^{4} + 4 {x}^{3} - 12 {x}^{2} + 5$

find the first two derivatives

${f}^{'} = 12 {x}^{3} + 12 {x}^{2} - 24 x$
${f}^{' '} = 36 {x}^{2} + 24 x - 24$

Set first derivative equal to zero

${f}^{'} = 0 \implies 12 {x}^{3} + 12 {x}^{2} - 24 x = 0$

Find the values of $x$

$12 x \left({x}^{2} + x - 2\right)$
$12 x \left({x}^{2} + 2 x - x - 2\right)$
$12 x \left[x \left(x + 2\right) - 1 \left(x + 2\right)\right]$
$12 x \left(x - 1\right) \left(x + 2\right)$

$12 x = 0$
$x = 0$

$x - 1 = 0$
$x = 1$

$x + 2 = 0$
$x = 2$

$x$ has three values

At $x = 0$

${f}^{' '} = 36 {\left(0\right)}^{2} + 24 \left(0\right) - 24 = - 24 < 0$

At x=0; f^'=0; f^('')<0

Hence the function has a maximum at $x = 0$

At $x = 1$

${f}^{' '} = 36 {\left(1\right)}^{2} + 24 \left(1\right) - 24$
${f}^{' '} = 36 + 24 - 24 = 36 > 0$

At x=0; f^'=0; f^('')<>0

Hence the function has a minimum at $x = 1$

At $x = 2$

${f}^{' '} = 36 {\left(2\right)}^{2} + 24 \left(2\right) - 24$
${f}^{' '} = 144 + 48 - 24 = 168 > 0$

At x=0; f^'=0; f^('')<>0

Hence the function has a minimum at $x = 2$

graph{3x^4+4x^3-12x^2+5 [-10, 10, -5, 5]} #