Solving Question 8: Differentiation and Analysis of Functions

In this article, we will walk through the step-by-step process of solving question 8, which involves differentiation and analyzing the given function. We will use techniques such as the sum rule, chain rule, and critical point analysis to understand the behavior of the function.

Step-by-Step Solution

Part 8i: Function and Its Derivative

The given function is:

F(x) 1/x 1/x2 x-1 x-2

To find the derivative of F(x), we use the power rule of differentiation, which states that:

For g(x) x^n, the derivative g'(x) nx^(n-1)

Applying this rule to our function:

F'(x) -1x^(-2) - 2x^(-3) -1/x2 - 2/x3

Part 8ii: Function Behavior Analysis

Now, let's analyze the behavior of the function F(x)

As x increases, the denominator in each term of F(x) increases. Since the function involves negative exponents, as the denominator increases, the fractional value decreases. Moreover, as the function is negative, the fractional value increases as x increases. For example:

This shows that as x increases, the function value decreases in magnitude but increases in absolute value.

Method to Solve the Problem

I. Finding the Derivative

To solve part 8i, we used the power rule as mentioned. The derivative of each term is calculated separately:

For F(x) 1/x, the derivative is F'(x) -1/x2

For F(x) 1/x2, the derivative is F'(x) -2/x3

Adding both results:

F'(x) -1/x2 - 2/x3

II. Identifying Critical Points

Next, we set the derivative F'(x) 0 to find the critical points:

-1/x2 - 2/x3 0

Multiplying through by x3

-x - 2 0

Solving for x gives:

x -2

Since x -2 is the only solution, this is the critical point.

III. Interval Analysis and Sign Testing

Next, we need to determine the intervals where the function is increasing or decreasing. We will divide the domain into intervals based on the critical point x -2 and test points within each interval:

Interval (-∞, -2): Test point x -3

Interval (-2, ∞): Test point x -1

Substitute these points into F'(x) to determine the sign:

For x -3: F'(-3) -1/(-3)2 - 2/(-3)3 -1/9 2/27 -1/27

For x -1: F'(-1) -1/(-1)2 - 2/(-1)3 -1 2 1

Based on these signs, we can conclude:

The function F(x) is decreasing in the interval [-∞, -2].

The function F(x) is increasing in the interval [-2, ∞].

IV. Finding Stationary Points

A stationary point is a point where the derivative of the function is zero. We already found the critical point x -2.

Substitute x -2 into the original function F(x) to find the corresponding y-value:

F(-2) 1/(-2) 1/(-2)2 -1/2 1/4 -1/4

Thus, the stationary point is:

(-2, -1/4)

Note that the question specifies that the critical point must be less than -1, and in this case, -2 is indeed less than -1.

Conclusion

In this article, we have solved question 8 by finding the derivative, analyzing the function's behavior, identifying the critical points, and determining the stationary point. Understanding these techniques is crucial for solving similar problems and optimizing search visibility in SEO.