Question

Identify a function \(f\) that has the given characteristics. Then sketch the function. $$ \begin{aligned} &f(-2)=f(4)=0 ; f^{\prime}(1)=0, f^{\prime}(x)<0 \\ &\text { for } x<1 ; f^{\prime}(x)>0 \text { for } x>1 \end{aligned} $$

Short answer

The function matching the characteristics given is \(f(x) = (x+2)(x-4)(x-1)\) and its sketch would show it passing through (-2, 0), (4, 0) with a turning point at (1, -3).

Step-by-step solution

Finding the roots and possible function

The function has roots at -2 and 4. This means \(f(x)\) is divisible by \((x - (-2))\) and \((x - 4)\). A possible function could then be \(f(x) = a(x+2)(x-4)\) where 'a' is a constant to be determined.

Identifying the possible function through derivative

When x<1, \(f'(x)<0\) signifying the function is decreasing, it then turns and increases for x>1. A turning point is characterized by the slope being 0 (\(f'(1) = 0\)). All this indicates that the function could be a cubic. Therefore, the function should look like \(f(x) = a(x+2)(x-4)(x-p)\), where p is the turning point at x=1. Hence, the potential function is \(f(x) = a(x+2)(x-4)(x-1)\).

Finding the constant 'a'

The constant 'a' can be any real number. However, to keep things simple and without loss of generality, we can choose \(a = 1\). Therefore, the function we are looking for is \(f(x) = (x+2)(x-4)(x-1)\).

Sketching the function

The function will have a positive coefficient for the leading x term (x^3) which means that the function will be upward facing. The roots that we found at -2 and 4 provide the x-intercepts of the graph, and the turning point at x=1 indicates that the function should have a local minimum at x=1. Sketch this function to complete the task. This can be done either manually or by using graphing software.

Key concepts

Finding Roots of a Function

Finding the roots of a function involves identifying the values for which the function equals zero. These roots, also known as zeros or x-intercepts, are critical for sketching the function's graph and understanding its behavior.

Roots are valuable in solving equations and inequalities because they represent the points where the graph of the function crosses or touches the x-axis. In the provided exercise, we concluded that the function has roots at -2 and 4. This means the function must be zero at these x-values. A simple cubic function that fits this requirement is of the form
\(f(x) = (x + 2)(x - 4)(x - p)\)
where \(p\) is another value that turns out to be a root, and in this case, it's determined through the analysis of the derivative to be 1, turning the equation into
\(f(x) = (x + 2)(x - 4)(x - 1)\).

Making the constant 'a' equal to 1 simplifies our search for a function that fits the given characteristics without affecting the shape or position of its graph relative to the x-axis.

Analyzing Function Derivatives

Analyzing the derivative of a function is essential for determining its increasing or decreasing behavior and identifying any turning points (local minima or maxima). The derivative of a function, denoted as \(f'(x)\), represents the slope of the tangent line to the graph at any given point.

Points where \(f'(x) = 0\) are potential turning points since this implies a horizontal tangent line, where the function switches from increasing to decreasing, or vice versa. In the original exercise, the derivative \(f'(x)\) is less than zero for \(x < 1\), indicating the function is decreasing on this interval. After x equals 1, \(f'(x)\) becomes greater than zero; hence, the function increases. The pivot at x=1 is where the derivative is zero, indicating a turning point.

This information allows us to deduce that the function has a local minimum at x=1 and judging by the derivative information, the function appears to be a cubic curve due to its single turning point and sign changes in the slope on either side of this point.

Graphing Cubic Functions

Graphing cubic functions, which are polynomials of degree three, is a vital skill to visualize the function's overall behavior and root locations. A cubic function generally has the form
\(f(x) = ax^3 + bx^2 + cx + d\),
where \(a, b, c,\) and \(d\) are constants, and 'a' determines the end behavior of the graph.

If 'a' is positive, as in our exercise, the cubic function will ultimately rise to positive infinity as x goes to positive infinity, and fall to negative infinity as x goes to negative infinity, resembling an 'S' shape. Our function
\(f(x) = (x+2)(x-4)(x-1)\)
has this exact characteristic, with 'a' being 1. A cubic graph passes through its roots, found in our case to be -2, 1, and 4. Between these roots lie the turns of the graph, which depend on the function's critical points.

As we sketched the function, we showed that the function's graph intersects the x-axis at the given roots and has a local minimum at x=1. This gives students a visual representation to go along with the algebraic analysis, aiding in complete comprehension of the function's attributes.