Question

In Exercises \(13-20\) , find all points of inflection of the function.3 $$y=x \sqrt{9-x^{2}}$$

Short answer

The points of inflection are \((- \sqrt3, 6\sqrt2),(\sqrt3, 6\sqrt2)\). The endpoints are not inflection points as the concavity does not change.

Step-by-step solution

Find the First Derivative

Firstly, find out the first derivative of the function \(y=x \sqrt{9-x^{2}}\). This is a product of two functions: \(u = x \) and \(v = \sqrt{9-x^{2}}\), so the product rule is applied.\n\nThe derivative of \(y\) with respect to \(x\) is: \n\n\(y' = u' \cdot v + u \cdot v'\)\n\nThe derivative of \(u(x)\) is \(u'=1\).\n\nThe derivative of \(v(x) = \sqrt{9-x^{2}}\) is \(v' = \frac{-x}{\sqrt{9-x^{2}}}\) using the chain rule.\n\nTherefore, the first derivative of \(y\) is given as:\n\n\(y'= 1 \cdot \sqrt{9-x^{2}} + x(-\frac{x}{\sqrt{9-x^{2}}})\)

Simplify the First Derivative

Simplify the first derivative for easier differentiation in the next step:\n\n\(y' = \sqrt{9-x^{2}} - \frac{x^{2}}{\sqrt{9-x^{2}}}\)

Find the Second Derivative

Now, find the second derivative of the function y which is the derivative of the first derivative. Apply the quotient rule which says that the derivative of \(\frac{u}{v}\) is \(\frac{u'v-uv'}{v^{2}}\). The second derivative will therefore be: \n\n\(y'' = v'(u'- v + uu' / v) = \frac{1 \cdot \sqrt{9-x^{2}} - x(-\frac{x}{\sqrt{9-x^{2}}})}{(\sqrt{9-x^{2}}) ^{2}} - \frac{x^{2}(-x/\sqrt{9-x^{2}} - \frac{1 \cdot \sqrt{9-x^{2}} - x(-\frac{x}{\sqrt{9-x^{2}}})}{(\sqrt{9-x^{2}}) ^{2}}}{(\sqrt{9-x^{2}})^{2}} \)

Simplify the Second Derivative

Simplify the second derivative:\n\n\(y'' = \frac{9-x^{2}}{(\sqrt{9-x^{2}}) ^{3}} - \frac{x^{2}}{(\sqrt{9-x^{2}}) ^{2}}\)

Determine the Points of Inflection

Points of inflection occur where the second derivative changes sign. So the task now is to solve \(y''=0\). Set \(y'' = \frac{9-x^{2}}{(\sqrt{9-x^{2}}) ^{3}} - \frac{x^{2}}{(\sqrt{9-x^{2}}) ^{2}} = 0\).\n\nUpon solving this you obtain \(x = -\sqrt3, \sqrt3\).! However, also remember to consider the end points, +3 and -3. These are the possible points of inflection.

Confirm the Points of Inflection

Check the intervalls around each point, whether y'' changes sign. If y'' changes sign around an x-value, (x, f(x)) is a point of inflection. If \(y'' > 0\), the function is concave upward. If \(y'' < 0\), the function is concave downward.

Key concepts

First Derivative

Understanding the first derivative is essential in calculus. It signifies the rate at which a function is changing at any given point and is crucial for determining critical points, which include maxima and minima of the function. The first derivative of a function is found using basic differentiation rules, the product rule, the chain rule, or the quotient rule, depending on the structure of the function.
To find the first derivative of the function given in the exercise, \( y = x \sqrt{9-x^{2}} \), we observe that it is a product of two simpler functions. The product rule is applied to efficiently find the derivative without expanding the equation, preserving the values where the function’s behavior changes, like at a point of inflection. Calculating the first derivative is the stepping stone to finding these inflection points.

Second Derivative

The second derivative provides us with information about the concavity of a graph, which tells us whether the graph is curving upwards or downwards. It is the derivative of the first derivative and can be used to locate points of inflection. These points are where the function transitions from curving up to curving down or vice versa.
In the exercise, finding the second derivative of the function \( y = x \sqrt{9-x^{2}} \) allows us to investigate the curvature and thus identify possible points of inflection. A change in the sign of the second derivative at a particular point indicates that it is a point of inflection, where the concavity of the function changes.

Product Rule

When differentiating a function that is the product of two or more functions, the product rule is a vital tool. It states that the derivative of a product \( u(x) \cdot v(x) \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). This rule is particularly useful when the functions cannot be easily multiplied to create a single expression.
For the function in the exercise, product rule simplifies the differentiation process by allowing us to treat each part of the product separately, which is especially helpful when the functions involved are complicated, as with the square root function.

Chain Rule

The chain rule enables us to differentiate composite functions. When a function \( u \) is nested inside another function \( v \), the derivative of the composite function \( v(u(x)) \) is \( v'(u(x)) \cdot u'(x) \). The chain rule is essential when functions are not simply x but expressions in terms of x.
When finding the first derivative for our function \( y = x \sqrt{9-x^{2}} \), the chain rule helps us differentiate \( \sqrt{9-x^{2}} \) as \( v = 9 - x^{2} \) and \( u = x^{2} \), leading us to \( u'(x) \) and \( v'(u(x)) \), which together give us the derivative of the nested function.

Quotient Rule

The quotient rule is used when we are differentiating a function that is the quotient of two other functions. It tells us that the derivative of \( \frac{u(x)}{v(x)} \) is given by \( \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{v(x)^2} \). This rule is particularly useful when the numerator and denominator cannot be simplified into a single term.
In finding the second derivative of our given function, we encounter a situation where we need to differentiate \( \frac{x^2}{\sqrt{9-x^2}} \) and the quotient rule provides an efficient method for doing so, ultimately enabling us to simplify the expression and find the points of inflection of the function.

Concavity

Concavity refers to the direction of the curve of the graph of a function. If the graph bends upwards like a cup, it’s said to be concave upward; if it bends downwards like an arch, it’s concave downward. The second derivative is a direct indicator of concavity.
In the context of the exercise, by determining where the second derivative is positive or negative, we can ascertain the concavity of the function at those intervals. Points of inflection are then identified by changes in concavity, which in turn are located by solving when the second derivative equals zero or does not exist. These points are significant because they describe transitions in the graph's curvature, which can affect the shape and behavior of the graph significantly.