Problem 40 - Single Variable Calculus

Question

Find the relative extrema of the function. $$ f(x)=3 \tan ^{-1} x-2 x $$

Step-by-Step Solution

Answer

The function \(f(x) = 3\tan^{-1}x - 2x\) has a relative maximum at \(x = \frac{1}{\sqrt{2}}\) and a relative minimum at \(x = -\frac{1}{\sqrt{2}}\).

Step 1: Find the derivative of the function

To find the critical points, we need to find the derivative of the function, \(f'(x)\). We will use the chain rule for the inverse tangent part and the constant rule for the linear part. For the inverse tangent, remember that \(\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}\), so applying the constant (3) and chain rule, we have \(3\frac{d}{dx}\tan^{-1}x = \frac{3}{1+x^2}\). For the linear part, the derivative of \(2x\) with respect to x is just \(2\). So, putting it all together, the derivative of the function is: $$ f'(x) = \frac{3}{1+x^2} - 2 $$

Step 2: Find the critical points

To find the critical points, we need to find the values of x for which \(f'(x) = 0\) or \(f'(x)\) is undefined: $$ \frac{3}{1+x^2} - 2 = 0 $$ Now, solve for x: $$ \frac{3}{1+x^2} = 2 \\ 3 = 2(1+x^2) \\ 3 = 2 + 2x^2 \\ 1 = 2x^2 \\ x^2 = \frac{1}{2} \\ x = \pm\frac{1}{\sqrt{2}} $$ So the critical points are \(x = \frac{1}{\sqrt{2}}\) and \(x = -\frac{1}{\sqrt{2}}\).

Step 3: Find the second derivative

To apply the second derivative test, we need to find the second derivative of the function, \(f''(x)\). Starting with the first derivative: $$ f'(x) = \frac{3}{1+x^2} - 2 $$ Then, find the second derivative of the given function: $$ f''(x) = \frac{d}{dx} \left( \frac{3}{1+x^2} - 2 \right) = -\frac{6x}{(1+x^2)^2} $$

Step 4: Apply the Second Derivative Test

Now, we will use the Second Derivative Test to classify the critical points: For \(x = \frac{1}{\sqrt{2}} \), we get: $$ f''\left(\frac{1}{\sqrt{2}}\right) = -\frac{6\frac{1}{\sqrt{2}}}{(1+\frac{1}{2})^2} < 0 $$ Since the second derivative is negative, we have a relative maximum at \(x = \frac{1}{\sqrt{2}}\). For \(x = -\frac{1}{\sqrt{2}} \), we get: $$ f''\left(-\frac{1}{\sqrt{2}}\right) = -\frac{-6\frac{1}{\sqrt{2}}}{(1+\frac{1}{2})^2} > 0 $$ Since the second derivative is positive, we have a relative minimum at \(x = -\frac{1}{\sqrt{2}}\). In conclusion, the function \(f(x) = 3\tan^{-1}x - 2x\) has a relative maximum at \(x = \frac{1}{\sqrt{2}}\) and a relative minimum at \(x = -\frac{1}{\sqrt{2}}\).

Key Concepts

Critical Points

To find where a function has relative extrema, we often start by identifying its critical points. Critical points occur where the derivative of the function equals zero or where the derivative is undefined.
For the function given, \(f(x) = 3 \tan^{-1} x - 2x\), we first find the derivative \(f'(x)\). This is \(f'(x) = \frac{3}{1+x^2} - 2\).

• Solve \(f'(x) = 0\) to find points where the slope of the tangent is flat.
• Check any points where \(f'(x)\) might be undefined, but in this case, \(f'(x)\) is always defined for all real numbers.

When solved, \(f'(x) = 0\) yields critical points at \(x = \pm \frac{1}{\sqrt{2}}\). These values of \(x\) identify where the function might go from increasing to decreasing or vice versa, indicating potential relative maxima or minima.

Second Derivative Test

The Second Derivative Test helps determine the nature of critical points. It involves taking the second derivative of the function and evaluating it at each critical point.
For our function \(f(x)\), the second derivative is \(f''(x) = -\frac{6x}{(1+x^2)^2}\).

• If \(f''(x) > 0\), the function is concave up, indicating a relative minimum.
• If \(f''(x) < 0\), the function is concave down, indicating a relative maximum.

For \(x = \frac{1}{\sqrt{2}}\), \(f''(x) < 0\), confirming a relative maximum. Meanwhile, for \(x = -\frac{1}{\sqrt{2}}\), \(f''(x) > 0\), confirming a relative minimum. Thus, the second derivative test effectively classifies the nature of these critical points.

Derivative of Inverse Tangent

The derivative of an inverse trigonometric function can sometimes be tricky. For the inverse tangent function, \(\tan^{-1}(x)\), the derivative is \(\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}\). Understanding this allows us to apply the chain rule effectively when differentiating functions that involve \(\tan^{-1}x\).
In our exercise, we found the derivative of \(3\tan^{-1}x\) by multiplying the constant 3 by the derivative of \(\tan^{-1}x\), resulting in \(\frac{3}{1+x^2}\).

• Applying the derivative rule for \(\tan^{-1}x\) is useful for finding the slope at any point \(x\).
• Knowing common derivatives like \(\tan^{-1}x\) helps streamline the process of finding critical points.

Mastering these derivative rules aids in tackling problems involving relative extrema and can be an essential step in your calculus toolkit.