Problem 56 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the inflection point(s), if any, of each function. $$ f(x)=\sqrt[5]{x} $$

Step-by-Step Solution

Answer

The inflection point of the function \(f(x) = \sqrt[5]{x}\) is at \((0, 0)\), where the second derivative changes its sign.

Step 1: Find the first derivative of the function

To differentiate the function \(f(x) = \sqrt[5]{x}\), we can rewrite the function as \(f(x) = x^{1/5}\) and apply the power rule: \[\frac{d}{dx}(x^n) = nx^{n-1}\]Thus, the first derivative of the function is: \[f'(x) = \frac{1}{5} x^{\frac{1}{5} - 1} = \frac{1}{5} x^{-\frac{4}{5}}\]

Step 2: Find the second derivative of the function

After finding the first derivative, we will now differentiate it to find the second derivative. Apply the power rule again: \[\frac{d}{dx} \left(\frac{1}{5}x^{-\frac{4}{5}}\right) = -\frac{4}{25} x^{-\frac{9}{5}}\]Thus, the second derivative of the function is: \[f''(x) = -\frac{4}{25} x^{-\frac{9}{5}}\]

Step 3: Determine critical points of the second derivative

To find the critical points, we must find where the second derivative is equal to zero or undefined. In this case, the second derivative is a rational function with a power of \(-\frac{9}{5}\) in the denominator, which means the second derivative is undefined when \(x=0\). We don't need to check for any zeros, as this rational function does not have any.

Step 4: Confirm the sign change of the second derivative

To confirm that the inflection point exists at \(x = 0\), we need to check if the sign of the second derivative changes. As x approaches 0 from the left, the second derivative becomes: \[f''(x) \approx -\frac{4}{25} x^{-\frac{9}{5}} \rightarrow +\infty\]As x approaches 0 from the right, the second derivative becomes: \[f''(x) \approx -\frac{4}{25} x^{-\frac{9}{5}} \rightarrow -\infty\]Thus, the second derivative changes its sign at \(x = 0\).

Step 5: Find the inflection point

Now that we have determined that there is an inflection point at \(x = 0\), we can find the corresponding point on the graph of the function: \[f(0) = \sqrt[5]{0} = 0\]Thus, the inflection point is \((0, 0)\).

Key Concepts

First Derivative

The concept of the first derivative is fundamental in calculus. It's like a compass, showing us the direction where a function is heading. When we take the first derivative of a function, we're essentially finding the function's slope at any given point. For the function in our exercise,

1. We start with: \(f(x) = \sqrt[5]{x}\) which can be rewritten as \(f(x) = x^{1/5}\).
2. Using the power rule, a handy tool that simplifies differentiation, we find the derivative: \(f'(x) = \frac{1}{5} x^{-4/5}\).

The first derivative can tell us about the critical points and where a function is increasing or decreasing, which are crucial insights when analyzing a graph's shape and direction. Remember, anywhere the first derivative is zero or undefined, we have critical points worth further investigation.

Second Derivative

The second derivative offers an even deeper look into how a function behaves. Where the first derivative tells us about slopes, the second derivative presents what happens to those slopes; it indicates concavity. In simpler terms, the second derivative tells us if the function bends upwards or downwards.For our original function, the second derivative is computed as:

• Starting from \(f'(x) = \frac{1}{5} x^{-4/5}\), we differentiate again using the power rule.
• This gives \(f''(x) = -\frac{4}{25} x^{-9/5}\).

An important part to notice here is where this derivative is undefined or zero. Values where it's undefined can indicate possible inflection points, which are significant in understanding where the graph itself changes curvature.

Critical Points

Critical points are the locations on a graph where interesting behavior occurs—they can be potential turning points or points where the function does something unpredictable, such as leveling off momentarily or changing direction. In the case of our exercise, critical points arise from examining our second derivative:

• We know that \(f''(x) = -\frac{4}{25} x^{-9/5}\) is undefined at \(x = 0\).
• This indicates something special could be happening at \(x = 0\) on the original function;
• It becomes a candidate for an inflection point if the concavity truly changes at that point.

Identifying critical points means examining where derivatives are zero or undefined, providing hints on where a function's graph might contain changes in its path or inflection.

Power Rule

The power rule is one of the first major differentiation tools you'll learn. It's straightforward and incredibly useful for finding derivatives of any polynomial function. It states:\[ \frac{d}{dx}(x^n) = nx^{n-1} \]In the context of our exercise, we used the power rule multiple times:

• First to differentiate \(f(x) = x^{1/5}\), leading to \(f'(x) = \frac{1}{5} x^{-4/5}\).
• Then again on \(f'(x)\) to find \(f''(x) = -\frac{4}{25} x^{-9/5}\).

By applying the power rule, we streamline the differentiation process, making it easier to handle functions with exponents. This technique shines as an efficient shortcut that becomes crucial when dealing with polynomial expressions in calculus.