Finding Tangent Line Of F(x) = 7x³ + 8x⁻⁵ + 3x At X = -1

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Determining the tangent line to a curve at a specific point is a fundamental concept in calculus, blending the ideas of derivatives and linear approximations. In this detailed exploration, we will walk through the process of finding the tangent line for the function f(x) = 7x³ + 8x⁻⁵ + 3x at the point where x = -1. This problem combines the power rule, negative exponents, and the point-slope form of a line, offering a comprehensive review of essential calculus techniques. Before we dive into the solution, let's briefly revisit the key concepts we'll be using: derivatives, tangent lines, and the point-slope form of a line.

Understanding Derivatives

At its core, the derivative of a function f(x) represents the instantaneous rate of change of the function with respect to its input variable, x. Geometrically, the derivative at a point gives the slope of the tangent line to the curve at that point. The derivative, denoted as f'(x), provides critical information about the function's behavior, including where it is increasing, decreasing, and has local extrema. Finding the derivative is the first crucial step in our problem, as it will give us the slope of the tangent line.

Tangent Lines: A Linear Approximation

A tangent line is a straight line that touches a curve at a single point, sharing the same slope as the curve at that point. It serves as a linear approximation of the function near that point. The equation of the tangent line is typically expressed in the slope-intercept form (y = mx + b) or the point-slope form (y - y₁ = m(x - x₁)), where m is the slope and (x₁, y₁) is the point of tangency. The concept of a tangent line is pivotal in calculus, providing insights into the local behavior of functions and serving as a foundation for more advanced topics.

The Point-Slope Form of a Line

The point-slope form is a versatile way to define the equation of a line when you know a point on the line and its slope. Given a point (x₁, y₁) and a slope m, the equation of the line is given by y - y₁ = m(x - x₁). This form is particularly useful when dealing with tangent lines, as we often know the point of tangency and the slope at that point (the derivative). We'll use the point-slope form to construct the equation of the tangent line once we've found the slope and the point on the curve.

Now that we've reviewed these essential concepts, let's tackle the problem step by step. Our first task is to find the derivative of the function f(x) = 7x³ + 8x⁻⁵ + 3x.

Step 1: Find the Derivative f'(x)

To find the derivative f'(x), we apply the power rule to each term in the function. The power rule states that if f(x) = axⁿ, then f'(x) = naxⁿ⁻¹. Applying this rule to each term in f(x) = 7x³ + 8x⁻⁵ + 3x, we get:

  • The derivative of 7x³ is 3 * 7x² = 21x².
  • The derivative of 8x⁻⁵ is -5 * 8x⁻⁶ = -40x⁻⁶.
  • The derivative of 3x is 1 * 3 = 3.

Combining these, the derivative f'(x) is:

f'(x) = 21x² - 40x⁻⁶ + 3

This derivative function gives us the slope of the tangent line at any point x on the curve. Our next step is to evaluate this derivative at x = -1 to find the slope of the tangent line at the specific point we're interested in.

Step 2: Evaluate f'(-1) to Find the Slope

Now that we have the derivative f'(x) = 21x² - 40x⁻⁶ + 3, we need to find the slope of the tangent line at x = -1. We do this by substituting x = -1 into the derivative:

f'(-1) = 21(-1)² - 40(-1)⁻⁶ + 3

Let's simplify this expression:

  • (-1)² = 1, so 21(-1)² = 21.
  • (-1)⁻⁶ = 1 / (-1)⁶ = 1 / 1 = 1, so -40(-1)⁻⁶ = -40.

Therefore,

f'(-1) = 21 - 40 + 3 = -16

So, the slope of the tangent line at x = -1 is -16. This value, -16, will be the m in our point-slope form equation. Now, we need to find the point on the curve where the tangent line touches, which means we need to find the y-coordinate corresponding to x = -1.

Step 3: Find the Point of Tangency

To find the point of tangency, we need to evaluate the original function f(x) = 7x³ + 8x⁻⁵ + 3x at x = -1. This will give us the y-coordinate of the point where the tangent line touches the curve:

f(-1) = 7(-1)³ + 8(-1)⁻⁵ + 3(-1)

Let's simplify:

  • (-1)³ = -1, so 7(-1)³ = -7.
  • (-1)⁻⁵ = 1 / (-1)⁵ = 1 / -1 = -1, so 8(-1)⁻⁵ = -8.
  • 3(-1) = -3.

Therefore,

f(-1) = -7 - 8 - 3 = -18

So, the point of tangency is (-1, -18). We now have all the information we need to write the equation of the tangent line: the slope m = -16 and the point of tangency (x₁, y₁) = (-1, -18). We'll use the point-slope form to construct the equation.

Step 4: Use the Point-Slope Form to Find the Tangent Line Equation

We have the slope m = -16 and the point (-1, -18). The point-slope form of a line is:

y - y₁ = m(x - x₁)

Substituting our values, we get:

y - (-18) = -16(x - (-1))

Simplifying,

y + 18 = -16(x + 1)

Now, let's distribute the -16 and solve for y to get the slope-intercept form:

y + 18 = -16x - 16

Subtract 18 from both sides:

y = -16x - 16 - 18

y = -16x - 34

Thus, the equation of the tangent line is y = -16x - 34. This matches option A from the given choices.

Final Answer

After meticulously working through each step, from finding the derivative to applying the point-slope form, we have determined that the tangent line for f(x) = 7x³ + 8x⁻⁵ + 3x at x = -1 is:

A. y = -16x - 34

This problem exemplifies the interplay between derivatives and linear approximations, providing a solid foundation for understanding more advanced calculus concepts. By breaking down the problem into smaller, manageable steps, we were able to confidently arrive at the correct solution. Understanding the derivative as the slope of the tangent line, the utility of the point-slope form, and the application of the power rule are all crucial skills reinforced by this exercise. This detailed walkthrough not only provides the answer but also serves as a guide to approaching similar problems in the future. Remember, the key to mastering calculus is practice and a solid grasp of fundamental concepts.

Keywords

  • Tangent Line
  • Derivative
  • Point-Slope Form
  • Power Rule
  • Calculus