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. $$ f(x)=-3 x \sqrt{x+1} $$

Short answer

The \(x\)-intercepts of the function are \(x = 0\) and \(x = -1\). The function's derivative equals 0 at \(x = -3/4\), which is indeed between 0 and -1.

Step-by-step solution

Find the x-intercepts

First, set the function \(f(x) = -3x\sqrt{x+1}\) equal to 0 and solve for x: \[0 = -3x\sqrt{x + 1}\] \[0 = x\sqrt{x + 1}\] \[0 = \sqrt{x^2 + x}\] The solutions to this equation are \(x = 0\) or \(x = -1\)

Differentiate the function

Next, we calculate the derivative of the function \(f(x) = -3x\sqrt{x + 1}\) using the chain rule: \[f^{\prime}(x) = -3\left[ \sqrt{x+1} + \frac{x}{2\sqrt{x+1}} \right]\]

Find the critical points

We set the derivative equation equal to 0 and solve for \(x\): \[0 = -3\left[ \sqrt{x+1} + \frac{x}{2\sqrt{x+1}} \right]\] Solving this equation, we find \(x = -3/4\)

Verify the condition

Finally, we verify that \(f^{\prime}(x) = 0\) for a point that lies between the \(x\)-intercepts found in Step 1. We have \(x\)-intercepts at \(x = -1\) and \(x = 0\) and a point where \(f^{\prime}(x) = 0\) at \(x = -3/4\). This lies between -1 and 0 so we can rightly conclude the condition is satisfied.

Key concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This field forms the basis for much of modern mathematics and is commonly used in science, engineering, and economics. Calculus provides tools, such as the derivative and the integral, which are used to evaluate the behavior of functions. For instance, when we discuss finding the x-intercepts of a function, calculus allows us to rigorously assess the points where a function crosses the x-axis. The key concept is to understand that the x-intercepts are the points where the function's value is zero. In calculus terms, solving the equation f(x) = 0 provides these intercepts.

When analyzing the behavior of functions, calculus also aids in understanding rates of change and accumulation processes, which are fundamental concepts exemplified by derivatives and integrals respectively.

Chain Rule

The Chain Rule is a fundamental theorem in calculus for differentiating composite functions. When we have a function that is composed of multiple other functions, as in the case of f(x) = -3x√(x+1), the chain rule allows us to find the derivative of it efficiently. Understanding the chain rule is critical for calculus students as it extends the ability to compute derivatives beyond elementary functions. The general form is (f(g(x)))' = f'(g(x))g'(x), meaning we take the derivative of the outer function applied to the inner function and multiply it by the derivative of the inner function.

Applying this to our example, the derivative of the outer function -3x with respect to x and the inner function √(x+1) with respect to x are both considered. Following the chain rule, we differentiate the square root function first, then multiply by the derivative of the inner function x+1, which leads us to find f'(x).

Critical Points

Critical points of a function are where the derivative is either zero or undefined, representing the points at which the graph of the function has a horizontal tangent line or a cusp. These points are important because they can indicate local maxima, minima, or inflection points of the function. In relation to the function given by f(x) = -3x√(x+1), finding its critical points involves calculating the derivative of the function and then determining where this derivative is equal to zero.

By setting f'(x) = 0, we're able to identify possible points of interest in the behavior of the function. In our exercise, solving the derivative equation for zero leads us to the critical point x = -3/4. This process is crucial in calculus because it allows us to analyze and predict the behavior of functions beyond just solving equations.

Derivative

The derivative is a measure of how a function changes as its input changes. It is a central concept in calculus that describes the slope of the tangent line to the function at any point. In practical terms, it can represent speed, acceleration, or other rates of change. When we calculated the derivative of the function f(x) = -3x√(x+1), we were finding a function f'(x) that provides the slope of f(x) at any point x.

In the context of the exercise, computing the derivative helps us find both the rate at which the function is changing and the critical points which could indicate features of interest - like local extrema or points of inflection. For example, the zero slope at x = -3/4 indicates a critical point where the function changes its concavity or has a peak or a trough.