Question

In Exercises, you are given \(f^{\prime}\). Find the intervals on which (a) \(f^{\prime}(x)\) is increasing or decreasing and (b) the graph of \(f\) is concave upward or concave downward. (c) Find the relative extrema and inflection points of \(f\). (d) Then sketch a graph of \(f\). $$ f(x)=3 x^{2}-2 $$

Short answer

Function \(f(x)\) is increasing for \(x > 0\) and decreasing for \(x < 0\). The function is concave upward for all \(x\). There are no inflection points or local extrema. The graph is a concave upward, upward-opening parabola with its vertex at \(x = 0\), \(y = -2\).

Step-by-step solution

Finding the derivative of \(f\)

Given \(f(x) = 3x^{2} - 2\), find the first derivative using the power rule of differentiation. The power rule states that \(d/dx[x^n]= nx^{n - 1}\), where \(n\) is a real number. Using this rule, the derivative of \(f(x)\) is \[f'(x) = 6x\].

Find the intervals where \(f'(x)\) is increasing or decreasing

A function is increasing at intervals where its derivative is positive and decreasing at intervals where its derivative is negative. The derivative \(f'(x) = 6x\) is zero at \(x = 0\). However, as the derivative is a linear function which only crosses the x-axis at this point and doesn't change its sign, it keeps increasing for all \(x\). So, \(f'(x)\) is decreasing for \(-\infty < x < 0\) and increasing for \(0 < x < \infty\).

Find the Concavity and Inflection Points

A function \(f(x)\) is concave upward where its second derivative \(f''(x)\) is positive and concave downward where \(f''(x)\) is negative. Any point where \(f''(x)\) changes its sign is an inflection point. The second derivative of \(f(x)\) is \(f''(x) = 6\), a constant. Therefore, it's always positive and the function is concave upward over the entire real line. Since \(f''(x)\) doesn't change its sign, there are no inflection points.

Find the relative extrema

A point is a local maximum if the derivative changes from positive to negative, or a local minimum if the derivative changes from negative to positive. In this case, \(f'(x) = 6x\) only equals zero at \(x = 0\). However, as the derivative doesn't change its sign, \(x = 0\) is neither a local maximum nor a local minimum. Therefore, \(f(x) = 3x^{2} - 2\) doesn't have any local extrema.

Sketch the graph

Based on the derivatives, we know the graph of \(f(x) = 3x^{2} - 2\) is a upward-opening parabola, with a minimum point (the vertex) at \(x = 0\), \(y = -2\). Moreover, the fact \(f'(x) = 6x\) tells us that the slope of the tangent line to the graph of \(f\) at any point is six times the x-coordinate of the point. This information, along with the knowledge that the graph of \(f\) is concave up everywhere, allows us to sketch at least a reasonably accurate graph.

Key concepts

Concavity and Inflection Points

Understanding concavity and inflection points is crucial in visualizing the shape of a graph. A section of a graph is \textbf{concave up} if it curves upward like a smiley face, while it is \textbf{concave down} if it curves downward like a frown. Mathematically speaking, we determine concavity by looking at the second derivative of a function.

If the second derivative, denoted as \( f''(x) \), is greater than zero on an interval, the function is concave up on that interval. Conversely, if \( f''(x) \) is less than zero, the function is concave down.

But what about \textbf{inflection points}? These are points on the graph where the concavity changes - the graph goes from curving upwards to curving downwards, or vice versa. To locate inflection points, we find where the second derivative is equal to zero or where it is undefined, and then check if there is a change in sign.

In the case of \( f(x) = 3x^{2} - 2 \), the second derivative is \( f''(x) = 6 \) and is always positive, meaning the graph of \( f(x) \) is always concave up. There are no inflection points since \( f''(x) \) does not change its sign.

Power Rule of Differentiation

The \textbf{power rule} is a fundamental technique in calculus for finding the derivative of functions involving powers of \( x \). The rule is straightforward: to differentiate \( x^n \), where \( n \) is a constant real number, you multiply by the exponent \( n \) and subtract one from the exponent.

The formula is expressed as \( \frac{d}{dx}[x^n] = nx^{n - 1} \). For instance, to find the first derivative of the function \( f(x) = 3x^{2} \) given in the problem, we apply the power rule to get \( f'(x) = 6x \).

This simple yet powerful technique enables us to understand how a function changes at any given point, providing the slope or rate of change at that point. Also, recognising that the slope of \( f(x) \) at \( x = 0 \) is zero helps us identify that this is where the vertex of the parabola is located.

Finding Extrema

Finding extrema (maximums and minimums) is a task that involves analyzing the behavior of the derivative of a function. The \textbf{first derivative test} can help us determine where a function has local extrema. These are points where the function reaches a local high or low relative to the surrounding points.

A function \( f(x) \) has a local \textbf{maximum} at point \( c \) if \( f'(c) \) changes from positive to negative as \( x \) increases through \( c \). On the other hand, it has a local \textbf{minimum} at point \( c \) if \( f'(c) \) changes from negative to positive.

However, in the given function \( f(x) = 3x^{2} - 2 \), since \( f'(x) = 6x \) does not change from positive to negative or vice versa at any point (because it only equals zero at \( x = 0 \)), this function has no local extrema. What this tells us is that the vertex of our parabola, though the lowest point on the graph, is not considered a local minimum because the parabola opens upwards without end.