Problem 15 - Calculus A New Horizon

Question

Find: (a) the intervals on which \(f\) is increasing, (b) the intervals on which \(f\) is decreasing, (c) the open intervals on which \(f\) is concave up. (d) the open intervals on which \(f\) is concave down, and (e) the \(x\) -coordinates of all inflection points. $$f(x)=\frac{x^{2}}{x^{2}+2}$$

Step-by-Step Solution

Answer

(a) Increasing on \((0, \infty)\), (b) Decreasing on \((-\infty, 0)\), (c) Concave up on \((-\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}})\), (d) Concave down on \((-\infty, -\sqrt{\frac{2}{3}})\) and \((\sqrt{\frac{2}{3}}, \infty)\), (e) Inflection points at \(x = \pm\sqrt{\frac{2}{3}}\).

Step 1: Find the First Derivative

To find where the function is increasing or decreasing, we need to differentiate the function first. Using the quotient rule, the first derivative of \(f(x)=\frac{x^{2}}{x^{2}+2}\) is given by:\[f'(x) = \frac{(2x)(x^2+2) - (x^2)(2x)}{(x^2+2)^2} = \frac{4x}{(x^2+2)^2}.\]

Step 2: Determine Critical Points

Set the first derivative equal to zero to find critical points:\[\frac{4x}{(x^2+2)^2} = 0.\]This occurs when the numerator is zero. Thus, \(4x = 0\) implies \(x = 0\). The denominator does not contribute extra zeros since it cannot be zero without making the expression undefined.

Step 3: Test Intervals for Increasing/Decreasing

Choose test points from intervals determined by the critical point \(x = 0\). Use points like \(-1\) and \(1\) to test the sign of \(f'(x)\):- For \(x < 0\), \(f'(-1) = \frac{-4}{(-1)^4 + 2} < 0\).- For \(x > 0\), \(f'(1) = \frac{4}{1^4 + 2} > 0\).Thus, \(f(x)\) is decreasing on \((-\infty, 0)\) and increasing on \((0, \infty)\).

Step 4: Find the Second Derivative

We need the second derivative to determine concavity. Differentiating \(f'(x)\) gives:\[f''(x) = \frac{d}{dx}\left(\frac{4x}{(x^2+2)^2}\right).\]Use the quotient rule again:\[f''(x) = \frac{(4)(x^2+2)^2 - (4x)(4x)(x^2+2)}{(x^2+2)^4} = \frac{4(x^2+2) - 16x^2}{(x^2+2)^3} = \frac{8 - 12x^2}{(x^2+2)^3}.\]

Step 5: Determine Intervals of Concave Up/Down

Set the second derivative equal to zero to find potential inflection points:\[\frac{8 - 12x^2}{(x^2+2)^3} = 0.\]This simplifies to \(8 - 12x^2 = 0\) or \(x^2 = \frac{2}{3}\), thus \(x = \pm\sqrt{\frac{2}{3}}\). Choose test points like \(-1, 0, 1\) in between these intervals:- For \(x < -\sqrt{\frac{2}{3}}\), \(f''(-1) < 0\), concave down.- For \(-\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}}\), \(f''(0) > 0\), concave up.- For \(x > \sqrt{\frac{2}{3}}\), \(f''(1) < 0\), concave down.

Step 6: Identify Inflection Points

Inflection points occur where the concavity changes sign, which are the points \(x = \pm\sqrt{\frac{2}{3}}\).

Key Concepts

Derivatives

In calculus, derivatives represent the rate at which a function changes at any given point, serving as a fundamental concept in understanding the behavior of functions. For a given function, such as \[f(x) = \frac{x^{2}}{x^{2}+2},\]we use derivatives to identify critical aspects like increasing or decreasing behavior.

To find the first derivative of this function, we apply the quotient rule. The quotient rule is used when differentiating a function that is the division of two other functions, and it is expressed as:\[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2},\]where \(u\) and \(v\) are functions of \(x\).

Applying this to our function, we get:\[f'(x) = \frac{4x}{(x^{2}+2)^{2}}.\]This derivative helps us investigate whether our function is increasing or decreasing.

Increasing and Decreasing Functions

Understanding when a function is increasing or decreasing is important for mapping its trend over an interval. The first derivative, \(f'(x)\), tells us precisely where this behavior occurs. Specifically:

• If \(f'(x) > 0\), the function is increasing on that interval.
• If \(f'(x) < 0\), the function is decreasing on that interval.
• A critical point occurs when \(f'(x) = 0\) or \(f'(x)\) is undefined.

With \(f'(x) = \frac{4x}{(x^2+2)^2}\), we find critical points by setting \(f'(x)=0\). This only happens when the numerator \(4x\) is zero, which gives us \(x = 0\).

Testing intervals around this critical point, such as \([-1, 0)\) and \((0, 1]\), shows that the function is:

• Decreasing on \((-\infty, 0)\)
• Increasing on \((0, \infty)\)

Concavity

Concavity relates to how the curvature of a function appears. It shows whether the function curves upwards or downwards on certain intervals. The second derivative, \(f''(x)\), provides insights into this behavior:

• If \(f''(x) > 0\), the function is concave up (cup-shaped) on that interval.
• If \(f''(x) < 0\), the function is concave down (cap-shaped) on that interval.

To identify concavity for our function, we compute the second derivative, which results in:\[f''(x) = \frac{8 - 12x^2}{(x^2+2)^3}.\]
By testing points in the intervals split by \(x = \pm\sqrt{\frac{2}{3}}\), it shows that:

• The function is concave down on \((-\infty, -\sqrt{\frac{2}{3}})\) and \((\sqrt{\frac{2}{3}}, \infty)\).
• It is concave up on \((-\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}})\).

Inflection Points

An inflection point occurs where a function changes its concavity. These points can be found where the second derivative changes sign. Our function's second derivative is:\[f''(x) = \frac{8 - 12x^2}{(x^2+2)^3}.\]
Setting \(f''(x) = 0\) helps locate these points. For\[8 - 12x^2 = 0\]the solution is \(x = \pm\sqrt{\frac{2}{3}}\).

Therefore, inflection points are at:

• \(x = -\sqrt{\frac{2}{3}}\)
• \(x = \sqrt{\frac{2}{3}}\)

At these points, the function changes from concave up to concave down or vice versa, making them crucial for understanding the overall shape of a function.