Problem 20 - Calculus for Biology and Medicine

Question

Determine all inflection points. $$ f(x)=(x-3)^{5}, 0 \in \mathbf{R} $$

Step-by-Step Solution

Answer

The function has an inflection point at \( x = 3 \).

Step 1: Understand Inflection Points

Inflection points occur where the second derivative of a function changes its sign. This means we need to find the second derivative of the function and solve for points where this sign change occurs.

Step 2: Find the First Derivative

The given function is \( f(x) = (x-3)^5 \). To find the first derivative, use the chain rule. The derivative of \((x-3)^5\) is \( f'(x) = 5(x-3)^4 \).

Step 3: Find the Second Derivative

Now, find the second derivative of the function. Differentiate \( f'(x) = 5(x-3)^4 \), which gives \( f''(x) = 20(x-3)^3 \).

Step 4: Solve for Inflection Points

Set the second derivative equal to zero to find potential inflection points: \( 20(x-3)^3 = 0 \). Solving this gives \( x = 3 \).

Step 5: Check Sign Change of the Second Derivative

To confirm \( x = 3 \) is an inflection point, check if the sign of \( f''(x) \) changes at \( x = 3 \). Consider values slightly less and greater than 3.

Step 6: Conclusion on Inflection Points

For \( x < 3 \), \( (x-3)^3 < 0 \) making \( f''(x) < 0 \) and for \( x > 3 \), \( (x-3)^3 > 0 \) making \( f''(x) > 0 \). The change of sign confirms an inflection point at \( x = 3 \).

Key Concepts

Second Derivative

The second derivative of a function is a valuable tool in calculus. It helps us understand the concavity of a function and whether it is curving upwards or downwards. To find the second derivative, we take the derivative of the first derivative. For the function given,

• Start with the first derivative: \( f'(x) = 5(x-3)^4 \).
• The second derivative is then found by differentiating this result.
• It results in \( f''(x) = 20(x-3)^3 \).

This second derivative tells us about the curvature of the function \( f(x) \). When the second derivative changes sign, it indicates a shift in the direction of the curvature which may signify an inflection point. This is crucial in identifying points where the graph changes from concave to convex or vice versa.

Sign Change

Recognizing a sign change in the second derivative is essential in identifying inflection points. An inflection point is a spot where the function's graph changes its concavity.

• Check the sign of \( f''(x) \) before and after any potential points.
• A change from positive to negative, or vice versa, indicates an inflection point.

For our function, \( f''(x) = 20(x-3)^3 \), setting this equal to zero yields \( x=3 \). By examining values just before and after 3 (like 2.9 and 3.1), we notice:

• For \( x < 3 \), \( f''(x) < 0 \).
• For \( x > 3 \), \( f''(x) > 0 \).

The swapping from negative to positive confirms a sign change, thus identifying \( x = 3 \) as a true inflection point.

Chain Rule

The chain rule is a powerful tool in differentiation used when dealing with composite functions, where one function is inside another. It simplifies the process of differentiation by breaking it down into manageable steps.

• For a composite function like \( (x-3)^5 \), think of it as an "outer" function and an "inner" function.
• The outer function is raised to the power, in this case \( ( ext{something})^5 \).
• The inner function is what's inside, \( x-3 \).

To apply the chain rule,

• Differentiates the outer function, treating the inner as a constant.
• Multiply by the derivative of the inner function.

This results in \( f'(x) = 5(x-3)^4 \) as the first derivative, obtained efficiently using the chain rule.

Differentiation

Differentiation is the process of finding the derivative of a function, which tells about rates of change or slopes of curves. It's a core concept in calculus and plays a significant role in finding the behavior of functions.

• Start with the original function, \( f(x) = (x-3)^5 \).
• Apply the differentiation techniques, like the chain rule, to find the derivative.
• Derivatives reveal insights into the function's rate of change, slope of tangent lines, and even more through higher-order derivatives.

In this problem, we begin by finding the first derivative \( f'(x) = 5(x-3)^4 \), and further differentiate it to get the second derivative \( f''(x) = 20(x-3)^3 \). Differentiation not only helps in calculating slope but also in understanding the curvature through second derivatives, leading to finding inflection points.