Question

In Exercises, find the point(s) of inflection of the graph of the function. $$ h(x)=(x-2)^{3}(x-1) $$

Short answer

The inflection point of the function \(h(x)=(x-2)^{3}(x-1)\) is at (2, 0).

Step-by-step solution

Find the first derivative

First, find the first derivative of the function \(h(x)=(x-2)^3(x-1)\) using the product rule, which states that the derivative of two functions multiplied together is the first function times the derivative of the second function plus the second function times the derivative of the first function, i.e. \((fg)' = f'g + fg'\). Apply this rule to obtain \(h'(x)\): \(h'(x)=3(x-2)^{2}(x-1)+(x-2)^{3}\)

Compute the second derivative

The next step is to find the second derivative of the function which involves taking the derivative of the first one obtained. Again, apply the product and chain rule: \(h''(x)=6(x-2)(x-1)+(3)\cdot2(x-2)^{2}+3(x-2)^{2}\). Simply it to get: \(h''(x)=6x^{2}-24x+36\)

Find potential points of inflection

To find the inflection points, you need to set the second derivative equal to zero and solve for x: \(6x^{2}-24x+36=0\). Simplify the quadratic equation: \((6x-12)^{2}=0\), which gives: \(x=2\)

Confirm the points of inflection

Take a number less than 2 and a number more than 2, plug them into the second derivative function, if the sign changes that means x=2 is a point of inflection. After plugging in the numbers, it can be observed that there is a change in sign. Therefore, the function h(x) does indeed have an inflection point at x = 2, and when plugged back into the original function, the coordinates of the point of inflection are (2, 0).

Key concepts

first derivative

The first derivative of a function is a fundamental tool used for determining where the function is increasing or decreasing, and it can also help in finding critical points. In our exercise, we are tasked with finding the first derivative of the function \(h(x) = (x-2)^3(x-1)\). To do this, we'll apply the product rule. This rule is essential when dealing with the derivative of a product of two functions, and its formula is:

• \((fg)' = f'g + fg'\)

Here, \(f = (x-2)^3\) and \(g = (x-1)\). The derivative \(h'(x)\) is found by computing:

• 3 times \((x-2)^2\) times \((x-1)\)
• Adding this to \((x-2)^3\) times the derivative of \((x-1)\), which is simply 1

Finally, you will arrive at:

• \(h'(x) = 3(x-2)^2(x-1) + (x-2)^3\)

This derivative helps understand the local behavior of the function around specific points.

second derivative

The second derivative provides insights into the curvature of the function and helps us find points of inflection, where the concavity of the function changes. To find the second derivative \(h''(x)\), we take the derivative of our first derivative, \(h'(x)\), applying the product and chain rules where necessary.
When calculating \(h''(x)\), start by differentiating each term in the first derivative. For this function, we continue using the product rule due to the product form of \((x-2)\) expressions. The chain rule becomes particularly handy when differentiating power terms like \((x-2)^2\).
After careful computation, we simplify \(h''(x)\) to \(6x^2 - 24x + 36\). This simpler quadratic form is much easier to work with when seeking potential inflection points by solving \(h''(x) = 0\).

product rule

The product rule is indispensable when differentiating products of functions. It can be remembered by the formula \((fg)' = f'g + fg'\). This tool allows us to find derivatives of functions like \(h(x) = (x-2)^3(x-1)\).
To leverage the product rule efficiently, we:

• Differentiate the first function \((x-2)^3\) to get \(3(x-2)^2\)
• Keep the second function \((x-1)\) unchanged, then repeat the process by switching roles and differentiate \((x-1)\), which results in 1

After these steps, the total contribution from each part is combined, resulting in an accurate first derivative that is essential for further analysis. It's useful to practice applying the product rule, as it appears frequently in calculus.

chain rule

The chain rule is a vital concept in calculus, allowing us to differentiate composite functions. The essence of the chain rule is in differentiating a function within another function, often noted with a formula like \((f(g(x)))' = f'(g(x))g'(x)\).
In our scenario, we use the chain rule alongside the power rule to deal with nested functions like \((x-2)^3\). The differentiation transforms as follows:

• Take the derivative of the outer function, resulting in \(3(x-2)^2\)
• Multiply it by the derivative of the inner function, which is 1 since it differentiates to \(x\)

The chain rule's flexibility allows us to tackle various derivative challenges, especially when combined with other rules like the product rule, offering a comprehensive toolkit for tackling complex differentiation tasks.