Let \(f(x)a\) Find And Simplify \(f(x)\) And \(f\)

Žŵğǁžƌŭηϰͳǁžƌƚśϯϭɖƚɛ1 Let 𝑓 𝑥a Find And Simplify 𝑓 𝑥 And 𝑓

Extracting the core instruction from the provided text, the assignment involves analyzing a function 𝑓(x). The goal is to find and simplify the given function 𝑓(x) and its derivative 𝑓'(x). Additionally, the task includes solving equations involving these functions and exploring their properties such as graphs, tangent lines, and critical points. Due to the fragmented and coded nature of the original prompt, the primary focus revolves around differentiation, solving for zeros, understanding slope behavior, and tangent line equations for a given function 𝑓(x).

Paper For Above instruction

In this paper, I will analyze the function 𝑓(x) by first determining its explicit form, simplifying it, and then exploring various calculus concepts including derivatives, critical points, tangent lines, and related rates as indicated by the fragmented prompts.

Understanding and Simplifying 𝑓(x)

The first step is to identify the exact form of 𝑓(x). Given the vague description, a typical approach involves assuming 𝑓(x) is a well-behaved function, such as a polynomial or a trigonometric function, based on the context of the related problems. For example, suppose 𝑓(x) = ax^2 + bx + c, a common quadratic function. Without a specific formula, suppose further that 𝑓(x) is provided or derived from a problem scenario.

Once the function is given, simplification involves algebraic reduction to its simplest form, removing common factors or combining like terms. For example, if 𝑓(x) = (x^2 - 4)/(x - 2), simplification would involve factoring numerator and denominator, then canceling common factors to obtain 𝑓(x) = x + 2 for x ≠ 2.

Deriving and Simplifying 𝑓'(x)

The next step entails differentiating 𝑓(x) to find 𝑓'(x), which indicates the slope of the tangent to the graph at any point x. Differentiation rules such as the power rule, quotient rule, or trigonometric derivatives may be employed depending on the form of 𝑓(x). Continuing the previous example, if 𝑓(x) = (x^2 - 4)/(x - 2), then 𝑓'(x) would be derived using the quotient rule:

𝑓'(x) = [(2x)(x - 2) - (x^2 - 4)(1)] / (x - 2)^2

which simplifies further to a rational function suitable for analysis of critical points and tangent lines.

Solving Equations and Critical Points

In the context of solving equations such as 𝑓(x) = 0 and 𝑓'(x) = 0, the goal is to find the x-values where the function intersects the x-axis and where its slope is zero, indicating potential maxima, minima, or inflection points. This involves setting the simplified derivative to zero and solving for x, which yields critical points. Corresponding y-coordinates are then obtained by evaluating 𝑓(x) at these x-values.

For example, if 𝑓(x) = x^3 - 3x + 1, then 𝑓'(x) = 3x^2 - 3. Setting 𝑓'(x) = 0 gives 3x^2 - 3 = 0 → x^2 = 1 → x = ±1. Substituting back into 𝑓(x) reveals the y-coordinates where the slope is zero, and the nature of these critical points can be analyzed via the second derivative test.

Secant Line at x=2

Finding the equation of the secant line to 𝑓(x) at x=2 requires two points on the graph: the point of interest (2, 𝑓(2)) and another point x = a nearby point. The slope of the secant line is given by:

m = (𝑓(a) - 𝑓(2)) / (a - 2)

Then, the equation of the secant line uses point-slope form:

y - 𝑓(2) = m(x - 2)

Choosing a small value of a close to 2, or approaching the point as a limit, aids in understanding the function's behavior near x=2.

Finding Points with Horizontal Tangents

Horizontal tangents occur where 𝑓'(x) = 0. After finding 𝑓'(x), solving 𝑓'(x) = 0 yields critical points where the slope of the tangent line is zero. Substituting these x-values into 𝑓(x) provides the y-coordinates at these tangent points. For example, with 𝑓(x) = sin(2x) + 2 sin(x), differentiation yields critical points at specific x-values where the derivative equals zero, leading to multiple points with horizontal tangents. Analytical solving of 𝑓'(x) = 0 leads to the x-coordinates, and then to the corresponding y-coordinates.

Analyzing Specific Trigonometric Functions

For functions involving sine or cosine, such as 𝑓(x) = sin(2x) + 2 sin(x), derivatives involve applying chain rule and sum rule, resulting in expressions like:

𝑓'(x) = 2 cos(2x) + 2 cos(x)

Setting 𝑓'(x) = 0 to find horizontal tangent points leads to solving trigonometric equations like:

2 cos(2x) + 2 cos(x) = 0 → cos(2x) + cos(x) = 0

which simplifies further using identities such as cos(2x) = 2 cos^2 x - 1, enabling solving for x and then evaluating 𝑓(x) at those points.

Finding x-intercept of a Tangent Line

Given a tangent line at a specific point, for example at x=π/6 on 𝑓(x) = sec^2(2x), the line's equation uses point-slope form combined with the derivative at that point. The x-intercept is obtained by setting y=0 in the tangent line equation and solving for x, which involves algebraic manipulation and trigonometric identities. Calculating the derivative of 𝑓(x) = sec^2(2x) at x=π/6 provides the slope, and the tangent line equation is constructed accordingly.

Extended Word Problems and Related Rates

The complex projectile and shoreline problems involve applying multivariable calculus concepts such as related rates, optimization, and trigonometry. For the projectile, the initial velocity, maximum height, and time to ground are derived by modeling the motion equations and using derivatives to connect rates of change. Similarly, for the shoreline problem, the minimum rowing speed involves optimization using derivatives to minimize travel time, considering the woman’s running and rowing velocities, distances, and angles involved.

Conclusion

This comprehensive analysis demonstrates the application of calculus principles — differentiation, solving equations, tangent line equations, and optimization — to understand the behavior of various functions, solve complex real-world problems, and interpret their graphical and algebraic properties.

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