Problem 7 - Essential Calculus

Question

Find the two \(x\) -intercepts of the function \(f\) and show that \(f^{\prime}(x)=0\) at some point between the two \(x\) -intercepts. $$ f(x)=x \sqrt{x+4} $$

Step-by-Step Solution

Answer

The function \(f(x)=x \sqrt{x+4}\) has \(x\)-intercepts at \(x_1 = 0\) and \(x_2 = -4\). \(f'(x)\) equals zero at \(x = -2\), which lies between \(x = 0\) and \(x = -4\)

Step 1: Find the x-intercepts

Start by setting \(f(x)\) equal to zero and solving for \(x\).\n0 = \(x \sqrt{x+4}\)\nThis implies that \(x = 0\) or \(x + 4 = 0\) which leads to two \(x\)-intercepts, \(x_1 = 0\) and \(x_2 = -4\) respectively.

Step 2: Find the derivative

In order to find the derivative \(f'(x)\) of the function \(f(x) = x \sqrt{x+4}\), we can apply the product rule which says the derivative of a product of two functions is the derivative of the first times the second plus the first times the derivative of the second. Thus we have \n \(f'(x) = \sqrt{x+4} + \frac{x}{2\sqrt{x+4}}\).

Step 3: Find where the derivative is zero

Now we can find the \(x\)-value(s) where the derivative equals to zero which means solving for \(x\) in the equation \n\(\sqrt{x+4} + \frac{x}{2\sqrt{x+4}} = 0\)\nSolving this equation gives \(x = -2\). Now, checking if this point lies in between the two \(x\)-intercepts, we find that it does indeed fall within the \(x\)-interval of \([-4,0]\).

Key Concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It's typically divided into two main areas: differential calculus and integral calculus. Differential calculus concerns the rate at which quantities change, while integral calculus deals with the accumulation of quantities. In our exercise, calculus principles are applied to find the x-intercepts of a function and to determine where the derivative of the function equals zero, which is instrumental in understanding the behavior of the function across its domain.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable. They're fundamental in calculus and indicate the slope of the tangent line to the graph of a function at a given point. This concept is crucial when analyzing how a function behaves, as it can help locate maxima, minima, and points of inflection. In the context of our exercise, calculating the derivative of the function \( f(x) = x \sqrt{x+4} \) is a key step in finding at what point the rate of change of the function is zero, which aids in understanding where the graph of the function has horizontal tangents.

Product Rule

The product rule is a derivative rule used when differentiating products of two or more functions. According to the product rule, the derivative of the product \( fg \) is \( f'g + fg' \), where \( f \) and \( g \) are functions of \( x \), and \( f' \) and \( g' \) are their respective derivatives. This powerful tool allows us to tackle the derivative of our function \( f(x) = x \sqrt{x+4} \) efficiently. By treating \( x \) and \( \sqrt{x+4} \) as separate functions, the product rule provides a systematic approach to find the derivative, which is the first step in our quest to understand the function's critical points.

Function Analysis

Function analysis involves examining the properties of a function to understand its behavior across its domain. It includes studying limits, continuity, derivatives, and graph features such as intercepts, asymptotes, and intervals of increase or decrease. By analyzing these characteristics, we can build a comprehensive picture of how a function acts. In the given problem, we analyze the function \( f(x) = x \sqrt{x+4} \) by finding its x-intercepts and determining where its derivative equals zero. These steps provide insights into the function's roots and the points where its slope is flat (horizontal tangent), which are essential in sketching the graph and predicting its behavior.