Let's analyze the given function $f(x) = 2x^3 - 6x^2 - 18x + 1$ and use the provided graph description to identify the relative maximum and minimum points.

(a) Relative Maximum:

From the description:

• The graph rises, becomes horizontal (indicating a potential maximum), and then falls at $x = -1$.

Steps to Calculate the Relative Maximum:

1. Find the derivative of the function $f(x)$ to identify critical points:

   $f'(x) = \frac{d}{dx} \left( 2x^3 - 6x^2 - 18x + 1 \right)$ $f'(x) = 6x^2 - 12x - 18$

2. Set the derivative equal to zero to find critical points:

   $6x^2 - 12x - 18 = 0$ $x^2 - 2x - 3 = 0$ $(x - 3)(x + 1) = 0$ The critical points are $x = 3$ and $x = -1$.

3. Use the second derivative test to determine whether these critical points correspond to maxima or minima:

   $f''(x) = \frac{d}{dx}(6x^2 - 12x - 18) = 12x - 12$

   • At $x = -1$: f''(-1) = 12(-1) - 12 = -24 \quad \text{(Since f''(-1) < 0$,$x = -1 is a relative maximum.)}

4. Calculate the value of the function at $x = -1$ to find the relative maximum: $f(-1) = 2(-1)^3 - 6(-1)^2 - 18(-1) + 1$ $f(-1) = -2 - 6 + 18 + 1 = 11$

Answer:

A. The function $f$ has a relative maximum at $x = -1$, and the value of this relative maximum is 11.

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(b) Relative Minimum:

From the description:

• The graph falls, becomes horizontal (indicating a potential minimum), and then rises at $x = 3$.

Steps to Calculate the Relative Minimum:

1. Second derivative test at $x = 3$: f''(3) = 12(3) - 12 = 36 - 12 = 24 \quad \text{(Since f''(3) > 0$,$x = 3 is a relative minimum.)}

2. Calculate the value of the function at $x = 3$ to find the relative minimum: $f(3) = 2(3)^3 - 6(3)^2 - 18(3) + 1$ $f(3) = 2(27) - 6(9) - 54 + 1 = 54 - 54 - 54 + 1 = -53$

Answer:

The function $f$ has a relative minimum at $x = 3$, and the value of this relative minimum is $-53$.

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Do you have any further questions, or would you like more details on this explanation?

Related Questions:

1. How can you use the first derivative test to identify relative maxima and minima?
2. What role does the second derivative play in determining the concavity of a function?
3. How can you find the inflection points of a cubic function?
4. What is the significance of the discriminant in determining the nature of critical points?
5. How would you sketch the graph of a cubic function based on its derivatives?

Tip:

For polynomial functions, critical points are found where the derivative equals zero. To classify these points, use the second derivative test: if positive, it's a minimum; if negative, it's a maximum.