Problem 51 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the inflection point(s), if any, of each function. $$ f(x)=6 x^{3}-18 x^{2}+12 x-15 $$

Step-by-Step Solution

Answer

There is an inflection point at (1, -15) for the function \(f(x) = 6x^3 - 18x^2 + 12x - 15\).

Step 1: Find the first derivative of the function

To find the critical points of the function \(f(x)\), we first find its first derivative \(f'(x)\) by applying differentiation rules: \(f(x) = 6x^3 - 18x^2 + 12x - 15\) \(f'(x) = \frac{d}{dx}(6x^3 - 18x^2 + 12x - 15)\) \(f'(x) = 18x^2 - 36x + 12\)

Step 2: Find the second derivative of the function

Now, let's find the second derivative, \(f''(x)\), by applying the differentiation rules again: \(f''(x) = \frac{d}{dx}(18x^2 - 36x + 12)\) \(f''(x) = 36x - 36\)

Step 3: Determine the critical points of the second derivative

To find the critical points, set the second derivative, \(f''(x)\), equal to zero: \(36x - 36 = 0\) Divide both sides by 36 to isolate x: \(x = 1\) So, there is only one critical point at \(x = 1\).

Step 4: Test for inflection point

An inflection point occurs where the function's second derivative changes signs. Let's test the intervals around the critical point: - Choose a test point to the left of \(x = 1\), say \(x = 0\): \(f''(0) = 36(0) - 36 = -36\) - Choose a test point to the right of \(x = 1\), say \(x = 2\): \(f''(2) = 36(2) - 36 = 36\) Since \(f''(x)\) changes sign from negative to positive around the critical point at \(x = 1\), this indicates there is an inflection point at \(x = 1\).

Step 5: Determine the inflection point coordinates

Now that we know there's an inflection point at \(x = 1\), let's find the corresponding y-value using the original function, \(f(x)\): \(f(1) = 6(1)^3 - 18(1)^2 + 12(1) - 15\) \(f(1) = 6 - 18 + 12 - 15\) \(f(1) = -15\) This means the function \(f(x)\) has an inflection point at the coordinates (1, -15).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which represents the rate of change of that function. In simpler terms,

• It shows how fast or slow a function's output changes with respect to its input.
• In our exercise, differentiation was crucial in deriving both the first and second derivatives of the polynomial function given.

For the function \(f(x) = 6x^3 - 18x^2 + 12x - 15\), applying differentiation rules gives us \(f'(x) = 18x^2 - 36x + 12\) for the first derivative.This derivative helps us locate critical points. Further differentiating, we find the second derivative \(f''(x) = 36x - 36\), which is vital for determining inflection points.Differentiation often involves basic calculus operations like the power rule, where we multiply by the original exponent and reduce it by one. Understanding these rules is essential in applying differentiation effectively.

Inflection Points

Inflection points are intriguing in calculus as they indicate a point on a curve where the concavity changes. This means the curve switches from bending upwards (concave up) to bending downwards (concave down), or vice versa.To spot inflection points, you check the second derivative of the function. In our task,

• The second derivative \(f''(x) = 36x - 36\) was set to zero to locate critical values.
• We solved for \(x\) and found that \(x = 1\) is a critical point.

Once you find potential inflection points, the next step is checking sign changes in the second derivative across these points.

• We tested on either side of \(x = 1\), with \(x = 0\) and \(x = 2\) to ensure there was a sign change.
• Since \(f''(x)\) changed from negative to positive, \(x = 1\) was confirmed as an inflection point.

Finally, to give the inflection point as a coordinate, we substitute \(x = 1\) back into the original function to derive \(f(1) = -15\), resulting in the point \((1, -15)\). This practice is crucial for graphical analysis and understanding the behavior of functions.

Critical Points

Critical points are found where the first derivative of a function equals zero or doesn't exist. They are essential in calculus as they can indicate points of local maxima, minima, or inflection points.In solving functions, as in our example,

• The first step was identifying critical points using the derivative \(f'(x)\).
• Setting \(f'(x) = 18x^2 - 36x + 12\) to zero to solve for \(x\).

After this step, you determine the nature of these critical points using the second derivative \(f''(x)\).

• If \(f''(x) > 0\) at a critical point, it's a local minimum.
• If \(f''(x) < 0\), it indicates a local maximum.

Lastly, if \(f''(x)\) changes signs, it could be an inflection point, like we concluded for \(x = 1\) in this problem. Recognizing these points allows us to describe a function's behavior and assists in creating a comprehensive sketch of its graph, which is a significant skill in calculus.