QuestionFind the relative maximums of the function $f(x) = (2x-1)(x+4)(x-2)$ .

Studdy Solution

STEP 1

What is this asking? Find the $x$ values where this curvy function $f(x)$ reaches its hilltops! Watch out! Don't mix up maximums and minimums, and make sure you find the relative maximums, not just the absolute highest point.
Also, remember that we're looking for the x values where these peaks occur, not the y values (the heights of the peaks).

STEP 2

1. Expand the function
2. Find the derivative
3. Find critical points
4. Determine the relative maximum

STEP 3

Alright, let's expand this function!
It's easier to take the derivative when it's not all factored like that.
We're doing this to make the next step (finding the derivative) much easier.

STEP 4

First, let's multiply $(2x-1)$ and $(x+4)$ : $(2x-1)(x+4) = 2x \cdot x + 2x \cdot 4 - 1 \cdot x - 1 \cdot 4 = 2x^2 + 8x - x - 4 = 2x^2 + 7x - 4$

STEP 5

Now, multiply the result by $(x-2)$ : \begin{align} (2x^2 + 7x - 4)(x-2) &= 2x^2 \cdot x + 7x \cdot x - 4 \cdot x - 2 \cdot 2x^2 - 2 \cdot 7x - 2 \cdot (-4) \\ &= 2x^3 + 7x^2 - 4x - 4x^2 - 14x + 8 \\ &= 2x^3 + 3x^2 - 18x + 8 \end{align} So, our expanded function is $f(x) = 2x^3 + 3x^2 - 18x + 8$ .

STEP 6

Now, let's find the derivative, $f'(x)$ , of our expanded function.
Remember, the derivative tells us the slope of the function at any point, and we're looking for where that slope is zero (at the hilltops).

STEP 7

Using the power rule, we get: $f'(x) = 3 \cdot 2x^{3-1} + 2 \cdot 3x^{2-1} - 1 \cdot 18x^{1-1} + 0 \cdot 8x^{0-1} = 6x^2 + 6x - 18$

STEP 8

To find the critical points, we set the derivative equal to zero and solve for $x$ .
These critical points are where the slope of our function is zero, meaning they could be maximums, minimums, or saddle points.

STEP 9

$6x^2 + 6x - 18 = 0$ We can divide both sides by $6$ to simplify: $x^2 + x - 3 = 0$

STEP 10

Now, we can use the quadratic formula to solve for $x$ : $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ In our case, $a=1$ , $b=1$ , and $c=-3$ , so: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-3)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2}$ So, our critical points are $x = \frac{-1 + \sqrt{13}}{2}$ and $x = \frac{-1 - \sqrt{13}}{2}$ .

STEP 11

We have two critical points.
Now, we need to figure out which one is a relative maximum.
We can do this by checking the sign of the second derivative, $f''(x)$ , at each critical point.

STEP 12

First, let's find the second derivative: $f''(x) = 2 \cdot 6x^{2-1} + 1 \cdot 6x^{1-1} - 0 \cdot 18x^{0-1} = 12x + 6$

STEP 13

Now, let's evaluate $f''(x)$ at $x = \frac{-1 + \sqrt{13}}{2}$ : $f''\left(\frac{-1 + \sqrt{13}}{2}\right) = 12\left(\frac{-1 + \sqrt{13}}{2}\right) + 6 = 6(-1 + \sqrt{13}) + 6 = 6\sqrt{13} > 0$ Since $f''(x)$ is positive, this point is a relative minimum.

STEP 14

Now, let's evaluate $f''(x)$ at $x = \frac{-1 - \sqrt{13}}{2}$ : $f''\left(\frac{-1 - \sqrt{13}}{2}\right) = 12\left(\frac{-1 - \sqrt{13}}{2}\right) + 6 = 6(-1 - \sqrt{13}) + 6 = -6\sqrt{13} < 0$ Since $f''(x)$ is negative, this point is a relative maximum.

STEP 15

The relative maximum occurs at $x = \frac{-1 - \sqrt{13}}{2}$ .

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