Finding Inflection Points For F(x) = 12x² - 4x + 7 + Ln(x)

In calculus, inflection points mark significant shifts in the concavity of a function's graph. These points signal where the function transitions from curving upwards to curving downwards, or vice versa. Identifying inflection points is crucial in understanding the behavior of a function and its graphical representation. This article delves into the process of finding inflection points, focusing on the function f(x) = 12x² - 4x + 7 + ln(x). We will explore the necessary calculus techniques, including finding first and second derivatives, and demonstrate how to apply these concepts to determine potential inflection points. Furthermore, we will analyze the second derivative's sign to confirm whether these points are indeed inflection points. This comprehensive approach will provide a clear understanding of how to locate and verify inflection points for a given function. Understanding inflection points provides valuable insights into the function’s behavior, such as identifying intervals of concavity and guiding optimization problems. By mastering this concept, one can better analyze mathematical models and real-world applications where functions play a central role.

Understanding Inflection Points

To fully grasp the concept of inflection points, it's essential to understand the underlying principles of concavity and how it relates to the function's derivatives. The concavity of a function describes the direction of its curvature. A function is concave up if its graph curves upwards, resembling a smile, while it is concave down if its graph curves downwards, resembling a frown. Mathematically, concavity is determined by the sign of the second derivative of the function. If the second derivative, denoted as f''(x), is positive over an interval, the function is concave up in that interval. Conversely, if f''(x) is negative, the function is concave down. An inflection point occurs where the concavity of the function changes. This means that at an inflection point, the second derivative either equals zero or is undefined, and the sign of the second derivative changes around that point. Geometrically, this corresponds to the point on the graph where the curve transitions from bending upwards to bending downwards, or vice versa. The significance of inflection points extends beyond mere mathematical curiosity. In real-world applications, inflection points can represent critical transitions in various phenomena. For instance, in economics, an inflection point on a cost curve might indicate the point of diminishing returns. In physics, it could represent a change in acceleration. Therefore, understanding how to find and interpret inflection points is invaluable in many fields.

Step-by-Step Calculation for f(x) = 12x² - 4x + 7 + ln(x)

To find the inflection points for the function f(x) = 12x² - 4x + 7 + ln(x), we follow a structured approach involving differentiation and analysis of the second derivative. This process ensures that we identify all potential inflection points and verify their nature. The first step is to find the first derivative f'(x), which represents the rate of change of the function. Applying the power rule and the derivative of the natural logarithm, we differentiate f(x) term by term. The derivative of 12x² is 24x, the derivative of -4x is -4, the derivative of the constant 7 is 0, and the derivative of ln(x) is 1/x. Combining these, we get f'(x) = 24x - 4 + 1/x. Next, we find the second derivative f''(x), which represents the rate of change of the first derivative and indicates the concavity of the function. Differentiating f'(x), the derivative of 24x is 24, the derivative of -4 is 0, and the derivative of 1/x (which can be written as x⁻¹) is -1/x². Thus, f''(x) = 24 - 1/x². To find potential inflection points, we set the second derivative equal to zero and solve for x. This gives us the equation 24 - 1/x² = 0. Solving this equation involves adding 1/x² to both sides, resulting in 24 = 1/x². Multiplying both sides by x² gives 24x² = 1, and dividing by 24 yields x² = 1/24. Taking the square root of both sides, we find x = ±√(1/24). Simplifying the square root, we get x = ±1/√(24), which further simplifies to x = ±1/(2√6). Rationalizing the denominator, we multiply the numerator and denominator by √6, giving x = ±√6/12. Since the natural logarithm ln(x) is only defined for positive values of x, we discard the negative solution. Therefore, the only potential inflection point occurs at x = √6/12.

Calculating the Second Derivative

The meticulous calculation of the second derivative is a pivotal step in identifying potential inflection points. The second derivative, denoted as f''(x), provides essential information about the concavity of the function. To accurately determine the second derivative, one must apply the rules of differentiation correctly and handle each term with precision. For the function f(x) = 12x² - 4x + 7 + ln(x), we first found the first derivative f'(x) = 24x - 4 + 1/x. Now, to find the second derivative, we differentiate f'(x) term by term. The derivative of 24x with respect to x is simply 24. The derivative of the constant term -4 is 0, as constants do not change with respect to x. The derivative of 1/x, which can also be expressed as x⁻¹, requires the power rule. The power rule states that the derivative of xⁿ is nxⁿ⁻¹. Applying this rule to x⁻¹, we get (-1)x⁻², which simplifies to -1/x². Combining these results, the second derivative f''(x) is 24 - 1/x². This expression is crucial because it allows us to analyze the concavity of the original function. By setting f''(x) equal to zero and solving for x, we can find the points where the concavity might change, i.e., the potential inflection points. Furthermore, by examining the sign of f''(x) in different intervals, we can determine where the function is concave up (f''(x) > 0) and where it is concave down (f''(x) < 0). Therefore, the accurate computation of the second derivative is indispensable for understanding the behavior of the function and locating its inflection points.

Finding Potential Inflection Points

To pinpoint potential inflection points, we analyze where the second derivative, f''(x), equals zero or is undefined. This step is crucial because inflection points occur where the concavity of the function changes, which corresponds to the second derivative transitioning through zero or a point of discontinuity. For our function, the second derivative is f''(x) = 24 - 1/x². Setting this equal to zero, we get the equation 24 - 1/x² = 0. Solving this equation involves several algebraic steps. First, we add 1/x² to both sides, yielding 24 = 1/x². Next, we multiply both sides by x², which gives us 24x² = 1. Then, we divide both sides by 24, resulting in x² = 1/24. To find the values of x, we take the square root of both sides. This gives us x = ±√(1/24). Simplifying the square root, we get x = ±1/√(24), which can be further simplified to x = ±1/(2√6). To rationalize the denominator, we multiply the numerator and denominator by √6, resulting in x = ±√6/12. However, we must consider the domain of the original function, f(x) = 12x² - 4x + 7 + ln(x). The natural logarithm ln(x) is only defined for positive values of x. Therefore, we discard the negative solution, x = -√6/12. This leaves us with one potential inflection point at x = √6/12. It is important to note that finding potential inflection points is only the first part of the process. We must also verify that the concavity changes at these points, which we will do in the next step by analyzing the sign of the second derivative around x = √6/12. This thorough approach ensures that we correctly identify all inflection points of the function.

Verifying Inflection Points

After finding potential inflection points, the next crucial step is to verify that the concavity of the function actually changes at these points. This verification process involves analyzing the sign of the second derivative, f''(x), in the intervals around each potential inflection point. If the sign of f''(x) changes from positive to negative or from negative to positive, then we can confirm that an inflection point exists at that x-value. For our function, f(x) = 12x² - 4x + 7 + ln(x), we identified a potential inflection point at x = √6/12. To verify this, we need to examine the sign of f''(x) = 24 - 1/x² in the intervals to the left and to the right of x = √6/12. First, let's consider an x-value to the left of √6/12. Since √6/12 is approximately 0.204, we can choose a value like x = 0.1. Plugging this into f''(x), we get f''(0.1) = 24 - 1/(0.1)² = 24 - 1/0.01 = 24 - 100 = -76. Since f''(0.1) is negative, the function is concave down in the interval to the left of √6/12. Next, let's consider an x-value to the right of √6/12. We can choose a value like x = 0.3. Plugging this into f''(x), we get f''(0.3) = 24 - 1/(0.3)² = 24 - 1/0.09 ≈ 24 - 11.11 ≈ 12.89. Since f''(0.3) is positive, the function is concave up in the interval to the right of √6/12. Since the sign of f''(x) changes from negative to positive at x = √6/12, we can definitively confirm that an inflection point exists at this x-value. To find the corresponding y-coordinate of the inflection point, we plug x = √6/12 into the original function f(x). This gives us f(√6/12) = 12(√6/12)² - 4(√6/12) + 7 + ln(√6/12). Simplifying this expression, we get f(√6/12) = 12(6/144) - (4√6)/12 + 7 + ln(√6/12) = 1/2 - (√6)/3 + 7 + ln(√6/12). Approximating the values, we get f(√6/12) ≈ 0.5 - 0.816 + 7 + ln(0.204) ≈ 0.5 - 0.816 + 7 - 1.59 ≈ 5.094. Therefore, the inflection point in (x, f(x)) form is approximately (√6/12, 5.094) or (0.204, 5.094) when rounded to the nearest thousandth.

Expressing the Inflection Point in (x, f(x)) Form

To express the inflection point in (x, f(x)) form, we need to find both the x-coordinate and the corresponding y-coordinate. We have already determined the x-coordinate to be x = √6/12. Now, we need to find the y-coordinate by plugging this value into the original function, f(x) = 12x² - 4x + 7 + ln(x). Substituting x = √6/12 into the function, we get f(√6/12) = 12(√6/12)² - 4(√6/12) + 7 + ln(√6/12). Let's simplify this expression step by step. First, we square √6/12, which gives us (√6/12)² = 6/144 = 1/24. Multiplying this by 12, we get 12(1/24) = 1/2. Next, we have the term -4(√6/12), which simplifies to -(4√6)/12 = -(√6)/3. Then, we have the constant term 7. Finally, we have the natural logarithm term, ln(√6/12). So, the expression becomes f(√6/12) = 1/2 - (√6)/3 + 7 + ln(√6/12). Now, let's approximate these values to get a numerical result. We know that √6 ≈ 2.449, so (√6)/3 ≈ 2.449/3 ≈ 0.816. Thus, the expression becomes f(√6/12) ≈ 0.5 - 0.816 + 7 + ln(√6/12). To approximate ln(√6/12), we first find the value of √6/12. We have √6/12 ≈ 2.449/12 ≈ 0.204. Now, we need to find ln(0.204). Using a calculator, we find that ln(0.204) ≈ -1.59. So, the expression becomes f(√6/12) ≈ 0.5 - 0.816 + 7 - 1.59. Combining these values, we get f(√6/12) ≈ 5.094. Therefore, the inflection point in (x, f(x)) form is approximately (√6/12, 5.094). Rounding to the nearest thousandth, we have (0.204, 5.094). This is the point where the concavity of the function changes, and it is crucial for understanding the behavior of the function and its graphical representation.

Conclusion

In conclusion, finding inflection points involves a systematic approach that combines calculus techniques and careful analysis. For the function f(x) = 12x² - 4x + 7 + ln(x), we successfully identified and verified an inflection point by following a step-by-step process. We began by computing the first derivative, f'(x), and then the second derivative, f''(x). The second derivative, f''(x) = 24 - 1/x², is crucial as it provides information about the concavity of the function. By setting f''(x) = 0, we found potential inflection points. In this case, we obtained x = ±√6/12, but we discarded the negative solution due to the domain restriction imposed by the natural logarithm in the original function. This left us with x = √6/12 as a potential inflection point. To verify that an inflection point indeed exists at x = √6/12, we analyzed the sign of f''(x) in the intervals around this point. We found that f''(x) changes sign from negative to positive at x = √6/12, confirming that the concavity changes at this point. Finally, to express the inflection point in (x, f(x)) form, we substituted x = √6/12 into the original function f(x) and calculated the corresponding y-coordinate. This gave us the inflection point (√6/12, 5.094), or approximately (0.204, 5.094) when rounded to the nearest thousandth. Understanding inflection points is vital for a comprehensive analysis of functions. They represent points where the rate of change of the slope changes, providing insights into the function's behavior and its graphical representation. The techniques discussed in this article can be applied to a wide range of functions, making the concept of inflection points a valuable tool in calculus and its applications. By mastering this process, one can gain a deeper understanding of how functions behave and how they can be used to model real-world phenomena.