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The first derivative of a function is a powerful tool in calculus, mainly used to find critical points where the function's growth rate changes. For the function given, \[f(x) = 3x^4 - 4x^3 + 6\], the first derivative is found by differentiating, resulting in \[f'(x) = 12x^3 - 12x^2\]. Critical points occur where this derivative is zero or undefined, and since this is a polynomial, it is defined everywhere. Therefore, setting \[12x^3 - 12x^2 = 0\] will help us find these points. Factoring gives \[12x^2(x-1) = 0\], indicating critical points at \(x = 0\) and \(x = 1\). Understanding where a function increases or decreases is crucial:

• When the first derivative is positive, the function is increasing.
• When it is negative, the function is decreasing.

Analyzing these intervals around critical points helps determine their nature, which can be classified further with the help of a graph.

While the first derivative tells us about the slope of a function, the second derivative provides information on its curvature or concavity. For our function \[f(x) = 3x^4 - 4x^3 + 6\],the second derivative is computed by differentiating the first derivative: \[f''(x) = 36x^2 - 24x\]. This second derivative helps us pinpoint possible inflection points by setting \[36x^2 - 24x = 0\]. Factoring out \(12x\) gives \[12x(3x - 2) = 0\], indicating candidate points at \(x = 0\) and \(x = \frac{2}{3}\). Testing these points involves assessing the function's concavity:

• If the second derivative is positive, the graph is concave up, resembling a bowl shape.
• If negative, the graph is concave down, resembling an upside-down bowl.

This information offers insights into potential inflection points, where the concavity changes from up to down or vice versa.

Inflection points are points on a curve where the concavity changes. To validate such points, not only we look for where the second derivative equals zero, but also a sign change in their intervals. **For our exercise:**- The candidate inflection points were found at \(x = 0\) and \(x = \frac{2}{3}\).By testing intervals around these points: - For example, using a test point, for **x < 0**, such as \(x = -1\), the second derivative \(f''(-1) = 60\), indicating the function is concave up. - At \(x = 1/2\), we find \(f''(1/2) = -6\), showing concave down. - Moving to \(x = 2\), the second derivative is \(f''(2) = 72\), concave up again.The change in concavity at these points confirms the presence of inflection points. Recognizing such points helps describe the overall shape of the graph, which is essential for understanding more complex function behaviors.