Question

Find the point of maxima of \(\mathrm{f}(\mathrm{x})=\int_{0}^{x^{2}} \frac{\mathrm{t}^{2}-\mathrm{t}}{\mathrm{e}^{\mathrm{t}}+1} \mathrm{dt}\).

Short answer

Answer: The function has a maxima at point \(x=1\).

Step-by-step solution

Compute the derivative of the function

We can find the derivative of the function \(f(x)\) using the Fundamental theorem of calculus, as follows: $$ f'(x) = \frac{d}{dx} \left[\int_0^{x^2} \frac{t^2-t}{e^t+1} dt\right]. $$ By applying the chain rule, we have: $$ f'(x) = \frac{d}{dx}\left[x^2\right] \cdot \frac{d}{du}\left[\int_0^u\frac{t^2 - t}{e^t + 1}dt\right]\Bigg|_{u=x^2} = 2x\left(\frac{x^4-x^2}{e^{x^2}+1}\right). $$

Find the critical points

To find the critical points, we need to find the points where \(f'(x)=0\): $$ 2x\left(\frac{x^4-x^2}{e^{x^2}+1}\right)=0. $$ The equation is equal to zero either when \(2x=0\) or when \(\frac{x^4-x^2}{e^{x^2}+1}=0\). The only solution for \(2x=0\) is \(x=0\). For the equation \(\frac{x^4-x^2}{e^{x^2}+1}=0\), it is only equal to zero when the numerator is equal to zero. So, we have: $$ x^4-x^2=0 \implies x^2(x^2 - 1) = 0. $$ This has solutions \(x=0,\pm1\). We already found \(x=0\) from the previous analysis, so our critical points are \(x=-1,0,1\).

Determine the point of maxima using the second derivative

We will now find the second derivative of \(f(x)\) and use it to determine the nature of the critical points: $$ f''(x) =\frac{d}{dx} \left[2x\left(\frac{x^4-x^2}{e^{x^2}+1}\right) \right]. $$ Using the product rule and the quotient rule, we get: $$ f''(x) = 2\left(\frac{x^4-x^2}{e^{x^2}+1}\right)+2x\left(\frac{(4x^3-2x)(e^{x^2}+1)-(2x^5-2x^3)e^{x^2}}{(e^{x^2}+1)^2}\right). $$ Now, we will evaluate the second derivative at our critical points: - \(f''(-1)\): Since \(x=-1\), it yields a positive result; hence, it's a point of minima. - \(f''(0)\): Since the first term disappears, it yields a positive result; hence, it's a point of minima. - \(f''(1)\): Since \(x=1\), it yields a negative result; hence, it's a point of maxima. Therefore, we can conclude that the function \(f(x)\) has a maxima at point \(x=1\).

Key concepts

Integral Calculus

Integral calculus is a branch of calculus that deals with the accumulation of quantities, such as areas under curves or the total growth produced by a continuously running process. The integral of a function represents the area under the curve of a graph of that function. A common problem in integral calculus is to find the area bounded by a curve and the x-axis from one point to another. This involves calculating definite or indefinite integrals. Definite integrals have upper and lower limits and give a numerical value, while indefinite integrals are represented as functions, which when derived, result in the original function that was integrated.

In the context of our exercise, we are dealing with a definite integral where the upper limit is a function of x, specifically, the square of x. We are looking for the point at which this area is at a maximum, which involves both integral calculus for defining the function as an area under a curve and differential calculus to find maxima through the use of derivatives.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a principle that links the concept of the derivative of a function to the concept of an integral and in essence, connects differential calculus with integral calculus. The FTC states that if a function is continuous over an interval, then the integral of that function over that interval can be computed using its antiderivative. It has two main parts: the first part provides a way to compute definite integrals, and the second part asserts that the derivative of the integral of a function is the original function.

In our exercise, the FTC is applied to compute the derivative of the integral of a function. We find the derivative of the given function, which incorporates an integral within its definition. By doing this, we effectively 'unwrap' the area-under-the-curve interpretation of the integral and instead focus on the rate at which this area changes concerning x - which leads us to find the critical points and ultimately the maxima.

Critical Points

Critical points in calculus refer to points on a graph where a function's derivative is either zero or undefined. These points are significant because they are candidates for where a function may achieve its highest or lowest values, known as maxima or minima, or where inflection points might occur. To locate the critical points, we first find the derivative of the function and then solve for the values of the independent variable when the derivative is equal to zero or does not exist.

When solving the exercise, after computing the derivative of the function using the FTC, we find the critical points by setting the derivative equal to zero. In this case, we find that the function has critical points at x = -1, 0, and 1. These points are critical for further analysis, as they can indicate where the graph of the function changes its concavity or where local maxima or minima may be located.

Second Derivative Test

The Second Derivative Test is a method used in calculus to determine whether a critical point of a differentiable function is a local maximum or minimum. This test involves taking the second derivative of the function, which gives us information about its concavity. If the second derivative at a critical point is greater than zero, the point is a local minimum; if it is less than zero, the point is a local maximum; and if it is equal to zero, the test is inconclusive, and the point may be an inflection point.

In our exercise, after finding the critical points, we use the Second Derivative Test to analyze their nature. By substituting the critical points into the second derivative of the function, we find that the function has a local maximum at x = 1, confirmed by a negative result of the second derivative. This test is crucial for distinguishing between the critical points and determining which represent the maxima or minima we seek.