Problem 23 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Test the indicated points for point of inflection:- i. \(f(x)=x^{3}-5 x^{2}+3 x-5\) at \(x=1, \frac{5}{3}, 2\). ii. \(f(x)=x^{4}-12 x^{3}+48 x^{2}\) at \(x=1,2,3,4\). iii. \(f(x)=x+x^{\frac{5}{3}}-2\) at \(x=0\). iv. \(f(x)=x^{2}+x^{\frac{4}{3}}+1\) at \(x=0\). v. \(f(x)=x+x^{\frac{2}{3}}+4\) at \(x=0\). vi. \(f(x)=x+x^{\frac{3}{5}}-3\) at \(x=0\). vii. \(f(x)=\sin x+\frac{x^{3}}{3 !}-\frac{x^{5}}{5 !}\) at \(x=0\). viii. \(f(x)=e^{x}-\frac{x^{2}}{2}-\frac{x^{3}}{6}\) at \(x=0\). ix. \(\quad f(x)=\sin x, \quad x \geq 0\) \(=x-\frac{x^{3}}{6}, \quad x<0\) at \(x=0\). \\{Ans. 0 is point of inflection\\}

Step-by-Step Solution

Answer

0 is a point of inflection for functions f(x)= \(sin(x) + \frac{x^{3}}{6} - \frac{x^{5}}{120}\), f(x)= \(e^{x} - \frac{x^{2}}{2} - \frac{x^{3}}{6}\), and f(x) for \(x \geq 0: sin(x)\) and for \(x<0: x - \frac{x^{3}}{6}\). The given points are not points of inflection for the other functions as the second derivative doesn't change its sign at those points.

Step 1: i. Finding second derivative for f(x) = x^3 - 5x^2 + 3x - 5:

Find the first derivative of f(x): f'(x) = 3x^2 - 10x + 3 Find the second derivative of f(x): f''(x) = 6x - 10 Now, compute the values of the second derivative at x = 1, 5/3, and 2: f''(1) = -4 f''(5/3) = 0 f''(2) = 2 Since f''(x) changes its sign at x = 1 and x = 2, then x = 1 and x = 2 are points of inflection. x = 5/3 is not a point of inflection because the second derivative equals 0 at this point without a change in sign.

Step 2: ii. Finding second derivative for f(x) = x^4 - 12x^3 + 48x^2:

Find the first derivative of f(x): f'(x) = 4x^3 - 36x^2 + 96x Find the second derivative of f(x): f''(x) = 12x^2 - 72x + 96 Now, compute the values of the second derivative at x = 1, 2, 3, and 4: f''(1) = 36 f''(2) = 0 f''(3) = -36 f''(4) = -96 Since f''(x) does not change its sign at x = 1, 2, 3, and 4, none of them are points of inflection.

Step 3: iii. Finding second derivative for f(x) = x + x^(5/3) - 2:

Find the first derivative of f(x): f'(x) = 1 + \(\frac{5}{3}\)x^(2/3) Find the second derivative of f(x): f''(x) = \(\frac{10}{9}\)x^(-1/3) Since the second derivative is always positive, there is no point of inflection, including x = 0.

Step 4: iv. Finding second derivative for f(x) = x^2 + x^(4/3) + 1:

Find the first derivative of f(x): f'(x) = 2x + \(\frac{4}{3}\)x^(1/3) Find the second derivative of f(x): f''(x) = 2 + \(\frac{4}{9}\)x^(-2/3) Since the second derivative is always positive, there is no point of inflection, including x = 0.

Step 5: v. Finding second derivative for f(x) = x + x^(2/3) + 4:

Find the first derivative of f(x): f'(x) = 1 + \(\frac{2}{3}\)x^(-1/3) Find the second derivative of f(x): f''(x) = -\(\frac{2}{9}\)x^(-4/3) Since the second derivative is always negative, there is no point of inflection, including x = 0.

Step 6: vi. Finding second derivative for f(x) = x + x^(3/5) - 3:

Find the first derivative of f(x): f'(x) = 1 + \(\frac{3}{5}\)x^(-2/5) Find the second derivative of f(x): f''(x) = -\(\frac{6}{25}\)x^(-7/5) Since the second derivative is always negative, there is no point of inflection, including x = 0.

Step 7: vii. Finding second derivative for f(x) = sin(x) + x^3/6 - x^5/120:

Find the first derivative of f(x): f'(x) = cos(x) + x^2/2 - x^4/24 Find the second derivative of f(x): f''(x) = -sin(x) + x - x^3/6 Since the second derivative changes its sign at x = 0, it is a point of inflection.

Step 8: viii. Finding second derivative for f(x) = e^x - x^2/2 - x^3/6:

Find the first derivative of f(x): f'(x) = e^x - x - x^2/2 Find the second derivative of f(x): f''(x) = e^x - 1 - x Since the second derivative changes its sign at x = 0, it is a point of inflection.

Step 9: ix. Finding second derivative for f(x) = sin(x) (x>=0) and f(x) = x - x^3/6 (x<0) at x = 0:

The second derivative for sin(x) is -sin(x). At x = 0, f''(x) = 0. For the second part, the second derivative for f(x) = x - x^3/6 is: f''(x) = -x At x = 0, f''(x) = 0. Since the second derivative is zero for both parts of f(x), x = 0 could be a point of inflection if there is a change in concavity. From the second derivative, we can observe that the concavity changes at x = 0, and hence, x = 0 is a point of inflection.

Key Concepts

Second Derivative Test

The Second Derivative Test is a powerful tool in calculus used to determine the concavity of a function at a given point. It involves finding the second derivative of a function, denoted as \(f''(x)\), and analyzing its value at specific points. Here's how it works:

• If \(f''(x) > 0\) for an interval around \(x\), the function is concave up in that interval.
• If \(f''(x) < 0\) for an interval around \(x\), the function is concave down in that interval.
• If \(f''(x) = 0\), the point is a candidate for an inflection point, but we need to check if the sign of \(f''(x)\) changes around this point to confirm it.

A change in the sign of the second derivative around a point indicates a change in concavity, signifying the presence of an inflection point. For accurate conclusions, once \(f''(x)\) is zero, verify a sign change by evaluating \(f''(x)\) slightly to the left and right of the point in question.

Differential Calculus

Differential Calculus is a branch of mathematics focused on the concept of the derivative, which measures how a function changes as its input changes. The derivative represents the slope of the tangent line to the function's graph at any point. When studying functions, particularly when testing for points of inflection, we use the differentiation process to find the first and second derivatives.

• The first derivative, \(f'(x)\), gives the rate of change or the slope of the function.
• The second derivative, \(f''(x)\), provides information about the curvature or concavity, helping us understand more about the function's behavior.

Mastering differential calculus involves knowing how to efficiently derive these expressions and interpret the results. The precision in finding derivatives aids in determining the nature of turning points and concavity for better function analysis. Exploring exercises involving derivative calculations reinforces understanding and prepares students to innovate solutions for complex mathematical problems.

Concavity

Concavity describes the way a function curves. This concept is crucial in calculus because it provides insight into the behavior of a function beyond just increasing or decreasing trends. Understanding concavity helps predict future values and graph shapes.

To assess concavity, the second derivative \(f''(x)\) is essential:

• If \(f''(x) > 0\), the graph of the function is concave up, resembling a bowl shape, where a curve opens upwards.
• If \(f''(x) < 0\), the graph is concave down, similar to an upside-down bowl, where the curve opens downwards.
• If \(f''(x) = 0\), it may suggest potential points of inflection, especially if there's a subsequent sign change around the point.

Concavity not only affects the appearance of graphs but also impacts optimization problems and analysis of functions. By studying changes in concavity, learners can discover points where the function might have maximum or minimum values or where it transitions from accelerating growth to decelerating growth.