Find the critical points of the given function.g(x)=13x3−12x2−12xg(x)=\frac13x^3-\frac12x^2-12xg(x)=31x3−21x2−12x

x = 4 , x = − 3 x=4,x=-3 x = 4 , x = − 3

Find the critical points of the given function. g ( x ) = 1 3 x 3 − 1 2 x 2 − 12 x g(x)=\frac13x^3-\frac12x^2-12x g ( x ) = 3 1 x 3 − 2 1 x 2 − 12 x

In this problem, we're asked to find the critical points of the given function. The function that we're given here is g of x equals 1 third x cubed minus 1 half x squared minus 12 x. Now remember that critical points are going to be where our derivative, so in this case, g prime of x, is either equal to 0 or it does not exist. So we want to start here by just finding the derivative of our function. So finding the derivative here, g prime of x, since I'm working with a basic polynomial function, I'm just going to be using the power rule here. Now in using the power rule, when I multiply this 3, it's gonna be multiplying that 1 third. 3 times 1 third is just 1. So the derivative of 1 third x cubed is just going to be x squared. Then the same thing is going to happen for this one half x squared. When I use the power rule and multiply it by that 2, it's going to cancel with that existing one half, and I'm just gonna be left here with x. Then finally, the derivative of that minus 12 x is just going to be minus 12. So this is my derivative. Now remember, it's going to make your life easier if you see that something is factorable from this point. So I can see that looking at this function, I can think about a number that multiplies to negative 12 and adds to negative 1 in order to factor this. Because I know that this factors to x minus 4 times x plus 3. Now because I have this factored and I wanna find where this derivative is equal to 0, I can set each of these individual factors equal to 0. So if I have x minus 4 equals 0 and x plus 3 equals 0, if I solve each of these for x, I'm going to add 4 to both sides here, leaving me with x is equal to positive 4. And then subtracting 3 from both sides, that gives me x equals negative 3. This gives me my 2 critical points at x equals 4 and x equals negative 3. Now remember that our critical points also exist where our derivative does not exist, but because here we are working working with a basic polynomial, there are no points for which this derivative does not exist. So these are my only two critical points. Thanks for watching. I'll see you in the next one.

Find the critical points of the given function. g ( x ) = 1 3 x 3 − 1 2 x 2 − 12 x g(x)=\frac13x^3-\frac12x^2-12x g ( x ) = 3 1 x 3 − 2 1 x 2 − 12 x

x = 4 , x = − 3 x=4,x=-3 x = 4 , x = − 3

x = 0 , x = 12 x=0,x=12 x = 0 , x = 12

x = − 4 , x = 3 x=-4,x=3 x = − 4 , x = 3

x = − 6 , x = 2 x=-6,x=2 x = − 6 , x = 2

5. Graphical Applications of Derivatives

Finding Global Extrema

Calculus

5. Graphical Applications of Derivatives

Finding Global Extrema

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Multiple Choice

Find the critical points of the given function.
$g(x)=\frac13x^3-\frac12x^2-12x$

$x=0,x=12$

$x=-4,x=3$

$x=-6,x=2$

$x=4,x=-3$

Verified step by step guidance

To find the critical points of the function $ g(x) = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 12x $, we first need to find its derivative, $ g'(x) $.

Differentiate the function: $ g'(x) = \frac{d}{dx} \left( \frac{1}{3}x^3 - \frac{1}{2}x^2 - 12x \right) $. Use the power rule for differentiation.

The derivative is $ g'(x) = x^2 - x - 12 $.

Set the derivative equal to zero to find the critical points: $ x^2 - x - 12 = 0 $.

Solve the quadratic equation $ x^2 - x - 12 = 0 $ using the quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = -1 $, and $ c = -12 $.

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