91Ó°ÊÓ

A derivative represents the rate at which a function changes at any given point. It's like a slope of the tangent line at a specific point on a curve. In our exercise, we have the function \(f(x) = 3x^3 - 4x^2 - 12x + 17\). To find how this function behaves, we need to compute its first and second derivatives.
The first derivative is obtained by differentiating the original function: \(f'(x) = 9x^2 - 8x - 12\). This first derivative helps us find the slope and thus tells us where the function is increasing or decreasing.
To dig deeper, we also compute the second derivative: \(f''(x) = 18x - 8\). This derivative informs us about the concavity of the function, helping to identify where the graph curves upwards or downwards.

Critical points are where the first derivative is zero or undefined. They are potential points where the function could have local maxima or minima. For our function, we set the first derivative equal to zero and solve: \[9x^2 - 8x - 12 = 0\]
Solving using the quadratic formula, we get \(x = \frac{14}{9}\) and \(x = -\frac{2}{3}\). These values divide the function into intervals and help us understand its behavior at specific points. By testing the sign of the first derivative in these intervals, we determine whether the function is increasing or decreasing around these critical points.

Concavity refers to the direction a curve bends. If a function is concave up, it looks like a cup and opens upwards. On the other hand, concave down looks like an upside-down cup. To determine concavity, we use the second derivative. For the given function, we find \(f''(x) = 18x - 8\).
We set this second derivative to zero to find potential inflection points: \[18x - 8 = 0 \rightarrow x = \frac{4}{9} \]
Testing values around \(x = \frac{4}{9}\), if \(f''(x) > 0\), the graph is concave up. If \(f''(x) < 0\), the graph is concave down. For instance, at \(x = 0\), \(f''(0) = -8\), indicating concave down. At \(x = 1\), \(f''(1) = 10\), indicating concave up.

Intervals of increase occur where the first derivative \(f'(x) > 0\), and decrease where \(f'(x) < 0\). From the critical points found in our problem, we examine the sign of \(f'(x)\) in the intervals: \(( -\infty, -\frac{2}{3})\), \((-\frac{2}{3}, \frac{14}{9})\), and \((\frac{14}{9}, \infty)\).
By testing points within each interval: \[ \text{For} \, x < -\frac{2}{3}, \, f'(-1) = 5 \rightarrow \text{positive, meaning increasing} \]
\[ \text{For} \, -\frac{2}{3} < x < \frac{14}{9}, \, f'(0) = -12 \rightarrow \text{negative, meaning decreasing} \]
\[ \text{For} \, x > \frac{14}{9}, \, f'(2) = 8 \rightarrow \text{positive, meaning increasing} \]
Hence, the function increases on the intervals \(( -\infty, -\frac{2}{3})\) and \((\frac{14}{9}, \infty)\), and decreases on \((-\frac{2}{3}, \frac{14}{9})\).

Points of inflection are where the concavity of a function changes. These points show where the graph switches from concave up to concave down or vice versa. To find these points, we set the second derivative \(f''(x)\) to zero and solve: \[18x - 8 = 0 \rightarrow x = \frac{4}{9} \]
This \(x = \frac{4}{9}\) is a candidate for a point of inflection. By testing values around \(x = \frac{4}{9}\), we see how concavity changes: For \(x < \frac{4}{9}\) (e.g., \(x = 0\)), \(f''(0) = -8 \rightarrow \text{concave down}\). For \(x > \frac{4}{9}\) (e.g., \(x = 1\)), \(f''(1) = 10 \rightarrow \text{concave up}\). Hence, the function switches concavity at \(x = \frac{4}{9}\), confirming it as a point of inflection.