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 \(f\). $$ f^{\prime}(x)=2 x+5 $$

Short answer

The derivative \(f'(x) = 2x + 5\) increases over \((-\infty, +\infty)\). Function \(f\) is concave upward over \((-\infty, +\infty)\). The function has a relative minimum at \(x= -\frac{5}{2}\) with the minimum value of \(-\frac{25}{4} + C\), and it has no inflection points. The graph of \(f\) is a parabola opening upwards, which can be sketched accordingly.

Step-by-step solution

- Determine where \(f'(x)\) increases or decreases

The derivative \(f'(x) = 2x + 5\) is a linear function which is increasing for all \(x\), since the coefficient of \(x\) is positive. No intervals of decreasing \(f'(x)\) are present.

- Identify the concavity of \(f\)

The second derivative is obtained by differentiating \(f'(x)\) again which will give \(f''(x) = 2\). Since \(f''(x)\) is positive for all \(x\), the graph of \(f\) is concave upward on the interval \((-\infty, +\infty)\) . There are no points where the graph is concave downward.

- Find the relative extrema

Relative extrema occur where the derivative changes signs and hence, where \(f'(x) = 0\). Solving for \(x\) when \(f'(x) = 0\) gives \(x = -\frac{5}{2}\). Therefore there is a relative minimum at \(x= -\frac{5}{2}\). The actual minimum value is found by integrating the derivative function to find \(f(x)\) and substituting \(x = -\frac{5}{2}\). The integral of \(f'(x)\) is \(f(x) = x^2 + 5x + C\), leading to a minimum value of \(f\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^2 -5*\frac{5}{2} + C = -\frac{25}{4} + C\).

- Find the inflection points

Since \(f''(x)\) does not equal zero for any \(x\), there are no inflection points.

- Sketch the graph of \(f\)

Use all previously gathered information to sketch the graph. The function \(f\) has a relative minimum at \(x = -\frac{5}{2}\) and is concave upward everywhere. This describes a parabola opening upwards, with the vertex at \(x = -\frac{5}{2}\).

Key concepts

Increasing and Decreasing Functions

Understanding when a function is increasing or decreasing is a fundamental aspect of calculus, directly tied to the interpretation of its derivative. For any function described by an equation such as f'(x) = 2x + 5, we look at the sign of its first derivative to determine where the function is increasing or decreasing.

Since the derivative f'(x) is essentially the function's rate of change, a positive derivative indicates that the function is increasing. In our example, because the coefficient of x in the derivative is positive, the function increases for all values of x. There are no intervals where f'(x) decreases since the derivative is a straight line with a positive slope. This simple but profound insight helps sketch the function's behavior without knowing the exact form of f(x).

Concavity and Inflection Points

The concavity of a function tells us about the direction in which a function curves. It's determined by the sign of the second derivative, f''(x). If f''(x) is positive for all x, the graph of the function is shaped like a bowl facing upwards, or concave upward. Conversely, if f''(x) is negative for all x, the graph is concave downward, like an upside-down bowl.

For a linear derivative like f'(x) = 2x + 5, the second derivative is simply the constant 2. Since it is always positive, our function f is concave upward for all values of x. Inflection points occur where the concavity changes - but in this case, because the second derivative never changes sign, there are no inflection points for this function.

Sketching Graphs of Functions

To sketch the graph of a function, you'll need to put together all the information you've gathered from its derivatives. Starting with the direction of the rates of change (increasing or decreasing) given by the first derivative, and continuing with the concavity indicated by the second derivative, you get a clear picture of its shape.

In the given example, f'(x) = 2x + 5 shows the function is always increasing. Adding to that, f''(x) = 2 reveals that the graph of the function is always concave upward. Thus, we can conclude that f(x) is a parabola that opens upwards. When sketching, mark the vertex of the parabola, which corresponds to the minimum point - provided by the value where f'(x) = 0. In our case, it's at x = -5/2. Draw a smooth curve passing through this point, ensuring it remains concave upward throughout.

Relative Extrema

Relative extrema refer to the peaks and valleys of a function - that is, the local maximum and minimum points. To find them, we set f'(x) = 0 and solve for x. Points where the sign of the derivative changes are likely locations of these extrema.

If you find a point where the derivative changes from negative to positive, you've located a relative minimum; where it changes from positive to negative, a relative maximum. In the case of the function with f'(x) = 2x + 5, setting the derivative to zero gives us x = -5/2, indicating a relative minimum since the derivative is positive before and after this point. There are no relative maxima because the function only increases. Remember that these are 'relative' not 'absolute' extrema - meaning they're relative to the surrounding points on the graph.