How do you find the exact relative maximum and minimum of the polynomial function of #f(x)= 2x^3-3x^2-12x-1#?

Question

Christopher Agresta · Accepted Answer

We use the first derivative to find all the critical points, and the second derivative to sort them in to maxima, minima and inflection points.

There are two things that we know about relative maxima and minima. The first is that the slope (first derivative) will be zero, but this alone is not sufficient. There are cases, known as inflection points where the slope is zero but the function continues to either increase or decrease.

This is why the second piece of information is so important - the curvature (second derivative) must be non-zero. We can start by finding all of the critical points of our function and then eliminating the inflection points:

#0=(del f(x)) / (del x)=6x^2-6x-12#

We use the quadratic equation to find the #x# values of the critical points:

#x = (6 +- sqrt(6^2-4*6*(-12)))/(2*6) = 2, -1#

We then test these values in the second derivative:

#(del^2 f(x))/(del x^2)=12x-6#

#(del^2 f(2))/(del x^2) = 18, (del^2 f(-1))/(del x^2)=-18#

From this we determine that the critical point at #x=-1# is a relative minimum and the critical point at #x=2# is a relative maximum.

We can now use these #x# values in the original polynomial to find the coordinates of these points:

#f(2)=-21, f(-1)=6#

So the points are:

relative minimum at #(2,-21)# and
relative maximum at #(-1,6)#

graph{2x^3-3x^2-12x-1 [-5, 5, -25, 15]}

Charles Anctil · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = 2x^3 - 3x^2 - 12x - 1 $, follow these steps:

1. Compute the derivative of the function $ f'(x) $.
2. Set $ f'(x) = 0 $ and solve for $ x $ to find critical points.
3. Determine the second derivative $ f''(x) $.
4. Evaluate the second derivative at each critical point.
5. If $ f''(x) > 0 $ at a critical point, it corresponds to a relative minimum. If $ f''(x) < 0 $, it corresponds to a relative maximum.

Let's go through these steps:

1. $ f'(x) = 6x^2 - 6x - 12 $.
2. Set $ f'(x) = 0 $ and solve for $ x $:

$ 6x^2 - 6x - 12 = 0 $.

$ x^2 - x - 2 = 0 $.

$ (x - 2)(x + 1) = 0 $.

$ x = 2 $ or $ x = -1 $.

3. $ f''(x) = 12x - 6 $.
4. Evaluate $ f''(x) $ at each critical point:

At $ x = 2 $, $ f''(2) = 12(2) - 6 = 18 $ (positive), so it's a relative minimum.
At $ x = -1 $, $ f''(-1) = 12(-1) - 6 = -18 $ (negative), so it's a relative maximum.

Therefore, the exact relative minimum is at $ x = 2 $ and the exact relative maximum is at $ x = -1 $.

Luke Alvarez · Answer

undefined To find the relative extrema of the function $f(x) = 2x^3 - 3x^2 - 12x - 1$, you need to first find its critical points by taking the derivative, setting it equal to zero, and solving for $x$. Then, you can use the second derivative test to determine whether each critical point corresponds to a relative maximum, minimum, or neither. 1. Find the derivative of $f(x)$: $$f'(x) = 6x^2 - 6x - 12$$ 2. Set $f'(x)$ equal to zero and solve for $x$: $$6x^2 - 6x - 12 = 0$$ $$2x^2 - 2x - 4 = 0$$ $$x^2 - x - 2 = 0$$ $$(x - 2)(x + 1) = 0$$ $$x = 2, \quad x = -1$$ 3. Find the second derivative of $f(x)$: $$f''(x) = 12x - 6$$ 4. Evaluate $f''(x)$ at each critical point: $$f''(2) = 12(2) - 6 = 18 > 0$$ $$f''(-1) = 12(-1) - 6 = -18 < 0$$ Based on the second derivative test: - $x = 2$ corresponds to a relative minimum because $f''(2) > 0$. - $x = -1$ corresponds to a relative maximum because $f''(-1) < 0$. 5. Determine the y-coordinates of the relative extrema by substituting the critical points into the original function: $$f(2) = 2(2)^3 - 3(2)^2 - 12(2) - 1 = -25$$ $$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) - 1 = 6$$ Therefore, the exact relative maximum occurs at $(-1, 6)$ and the exact relative minimum occurs at $(2, -25)$.