Given the function $f(x) = 2x^3 - 6x + 3$, we need to find:

1. Intervals of increasing and decreasing
2. Local maximum and local minimum points
3. Intervals of concavity
4. Inflection points

Step 1: First derivative (for increasing/decreasing intervals and critical points)

To find where the function is increasing or decreasing, we first find the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(2x^3 - 6x + 3) = 6x^2 - 6$

Now, set $f'(x) = 0$ to find the critical points:

$6x^2 - 6 = 0$ $x^2 = 1$ $x = \pm 1$

So, the critical points are $x = 1$ and $x = -1$.

Step 2: Test for increasing/decreasing intervals

We use a sign chart for $f'(x) = 6x^2 - 6$ to determine where the function is increasing or decreasing.

• For $x < -1$, choose $x = -2$: $f'(-2) = 6(-2)^2 - 6 = 24 - 6 = 18 \quad (\text{positive, so increasing on } (-\infty, -1))$
• For $-1 < x < 1$, choose $x = 0$: $f'(0) = 6(0)^2 - 6 = -6 \quad (\text{negative, so decreasing on } (-1, 1))$
• For $x > 1$, choose $x = 2$: $f'(2) = 6(2)^2 - 6 = 24 - 6 = 18 \quad (\text{positive, so increasing on } (1, \infty))$

Step 3: Local maximum and minimum

• At $x = -1$, the function changes from increasing to decreasing, so there is a local maximum at $x = -1$.
• At $x = 1$, the function changes from decreasing to increasing, so there is a local minimum at $x = 1$.

To find the values of $f(x)$ at these points:

• $f(-1) = 2(-1)^3 - 6(-1) + 3 = -2 + 6 + 3 = 7$
• $f(1) = 2(1)^3 - 6(1) + 3 = 2 - 6 + 3 = -1$

So the local maximum is at $(-1, 7)$, and the local minimum is at $(1, -1)$.

Step 4: Second derivative (for concavity and inflection points)

To determine the concavity, we find the second derivative $f''(x)$:

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

Now, set $f''(x) = 0$ to find potential inflection points:

$12x = 0$ $x = 0$

So, $x = 0$ is a possible inflection point.

Step 5: Test for concavity and inflection points

We check the sign of $f''(x)$ on either side of $x = 0$:

• For $x < 0$, choose $x = -1$: $f''(-1) = 12(-1) = -12 \quad (\text{concave down on } (-\infty, 0))$
• For $x > 0$, choose $x = 1$: $f''(1) = 12(1) = 12 \quad (\text{concave up on } (0, \infty))$

Since the concavity changes at $x = 0$, there is an inflection point at $x = 0$. To find the corresponding $y$-value:

$f(0) = 2(0)^3 - 6(0) + 3 = 3$

So the inflection point is at $(0, 3)$.

Summary of results:

1. Intervals of increasing: $(-\infty, -1)$ and $(1, \infty)$
2. Intervals of decreasing: $(-1, 1)$
3. Local maximum: $(-1, 7)$
4. Local minimum: $(1, -1)$
5. Intervals of concavity:
   • Concave down on $(-\infty, 0)$
   • Concave up on $(0, \infty)$
6. Inflection point: $(0, 3)$

Would you like further details on any of these steps?

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Here are 5 related questions:

1. What happens if we change the coefficients in the cubic function?
2. How do you determine if a point is a local maximum or minimum without using the first derivative test?
3. Can a function have more than one inflection point?
4. How does the shape of a cubic function change with different coefficients?
5. How do you find critical points for more complex functions like trigonometric or logarithmic?

Tip: When analyzing concavity, always remember that inflection points occur where the concavity changes from up to down or vice versa.