Question

guided practice
determine the extrema of the following function.
$f(x)=3x^{4}-4x^{3}-24x^{2}+48x$
minima: (1, 23) maxima: (-2, -112), (2, 16)
minima: (-2, -112), (2, 16) maxima: (1, 23)
minima: (-2, -112) maxima: (1, 23), (2, 16)
minima: (2, 16) maxima: (1, 23), (-2, -112)
extrema are turning points on a graph.
when a graph goes from decreasing to increasing, the turning point is a minimum.
when a graph goes from increasing to decreasing, the turning point is a maximum.

Explanation:

Step1: Find the derivative

Differentiate $f(x)=3x^{4}-4x^{3}-24x^{2}+48x$ using power - rule. $f^\prime(x)=12x^{3}-12x^{2}-48x + 48$.

Step2: Set the derivative equal to zero

$12x^{3}-12x^{2}-48x + 48 = 0$. Factor out 12: $12(x^{3}-x^{2}-4x + 4)=0$. Then factor by grouping: $12[x^{2}(x - 1)-4(x - 1)]=0$, so $12(x - 1)(x^{2}-4)=0$. Further factor $x^{2}-4=(x + 2)(x - 2)$. The critical points are $x=-2,1,2$.

Step3: Use the second - derivative test

Differentiate $f^\prime(x)$ to get $f^{\prime\prime}(x)=36x^{2}-24x-48$.
For $x=-2$: $f^{\prime\prime}(-2)=36\times(-2)^{2}-24\times(-2)-48=144 + 48-48=144>0$, so $f(-2)=3\times(-2)^{4}-4\times(-2)^{3}-24\times(-2)^{2}+48\times(-2)=48 + 32-96 - 96=-112$ is a local minimum.
For $x = 1$: $f^{\prime\prime}(1)=36\times1^{2}-24\times1-48=36-24-48=-36<0$, so $f(1)=3\times1^{4}-4\times1^{3}-24\times1^{2}+48\times1=3-4-24 + 48=23$ is a local maximum.
For $x = 2$: $f^{\prime\prime}(2)=36\times2^{2}-24\times2-48=144-48-48=48>0$, so $f(2)=3\times2^{4}-4\times2^{3}-24\times2^{2}+48\times2=48-32-96 + 96=16$ is a local minimum.

Answer:

Minima: (-2, -112), (2, 16) Maxima: (1, 23)