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Chapter 3: Problem 54

Find an equation of the tangent line to the graph of \(f\) at \(P\). \(f(x)=3 x^{2}-2 \sqrt{x} ; \quad P(4,44)\)

Short Answer

Expert verified
The equation of the tangent line is \(y = 23.5x - 50\).

Step by step solution

01

Understand the Problem

We are asked to find the equation of the tangent line to the graph of the function \( f(x) = 3x^2 - 2\sqrt{x} \) at a given point \( P(4,44) \). The point \( P \) represents \( (x, f(x)) \) and is on the function's graph. Our goal is to use differentiation to find the slope of the tangent line at \( x = 4 \) and then develop the equation of the line using point-slope form.
02

Find the Derivative

First, we differentiate the function \( f(x) = 3x^2 - 2\sqrt{x} \) using basic differentiation rules. The derivative of \( f(x) \) is \( f'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(2\sqrt{x}) \). The derivative \( \frac{d}{dx}(3x^2) = 6x \), and for \( \frac{d}{dx}(2\sqrt{x}) = 2\frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}} \). Thus, \( f'(x) = 6x - \frac{1}{\sqrt{x}} \).
03

Evaluate the Derivative at the Given Point

To find the slope of the tangent line at \( x = 4 \), we substitute \( x = 4 \) into the derivative: \( f'(4) = 6(4) - \frac{1}{\sqrt{4}} = 24 - \frac{1}{2} = 23.5 \).Therefore, the slope of the tangent line at \( P(4,44) \) is 23.5.
04

Use Point-Slope Form to Find the Equation of the Tangent Line

The point-slope form equation of a line is given by \( y - y_1 = m(x - x_1) \) where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line.Since our point is \( P(4,44) \) and the slope is 23.5, we substitute these values into the equation: \( y - 44 = 23.5(x - 4) \).
05

Simplify the Equation

Now, simplify the equation of the line:Start with \( y - 44 = 23.5(x - 4) \).Distribute 23.5: \( y - 44 = 23.5x - 94 \).Add 44 to both sides to solve for \( y \): \( y = 23.5x - 50 \).This equation \( y = 23.5x - 50 \) represents the line tangent to \( f(x) \) at \( P(4,44) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is an essential concept in calculus that involves computing the derivative of a function. Derivatives help determine how a function changes at any given point. In simpler terms, the derivative is a tool used to find the slope of a function at a specific point, telling us how steep the curve is.
Let's break it down:
  • When you differentiate a function, you find a new function, called the derivative, that gives the slope of the original function at any point.
  • The process relies on rules like the power rule, product rule, and chain rule, among others.
  • For example, considering the function in our exercise, we have:
    - For the term \(3x^2\), differentiation gives us \(6x\) using the power rule where \( \frac{d}{dx}(x^n) = nx^{n-1} \).
  • For the term \(-2\sqrt{x}\), differentiation involves recognizing \(\sqrt{x}\) as \(x^{1/2}\). Using the power rule, \( \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} \), which simplifies to \( \frac{1}{2\sqrt{x}} \).
  • Putting it all together, the function \(3x^2 - 2\sqrt{x}\) has the derivative \(6x - \frac{1}{\sqrt{x}}\).
By finding the derivative, we can determine the rate of change or the slope at the point of interest.
Slope of Tangent
When we talk about the slope of a tangent line, we're referring to the rate at which the function changes at a specific point. The tangent line serves as a "best fitting" straight line that touches the curve at exactly one point, matching the curve's slope at that point.
To find this slope, we use the function's derivative:
  • First, evaluate the function's derivative at the given point. This involves substituting the x-coordinate from the point into the derivative.
  • In our exercise, the derivative \(f'(x) = 6x - \frac{1}{\sqrt{x}}\) was calculated. To find the slope at point \(P(4,44)\), substitute \(x = 4\).
  • The calculation is straightforward: \(f'(4) = 6 \times 4 - \frac{1}{\sqrt{4}} = 24 - 0.5 = 23.5\).
  • Therefore, the slope of the tangent line at \(P(4,44)\) is 23.5, indicating how the function behaves and changes at this point. It's steep, showing a strong rate of change.
Understanding the slope is crucial since it tells us about the behavior of the function at an exact point, letting us model real-life scenarios and make precise predictions.
Point-Slope Form
The point-slope form is a straightforward way to describe a line when you know a point on the line and the slope. The equation looks like this: \(y - y_1 = m(x - x_1)\), where:
  • \(m\) is the slope of the line.
  • \((x_1, y_1)\) is a point on the line.
This form provides a simple method to write the equation of a line, primarily using available data:
  • In our example, the point \(P(4,44)\) offers \(x_1 = 4\) and \(y_1 = 44\).
  • The slope \(m = 23.5\), calculated earlier.
  • Substituting these values into the point-slope formula, we have: \(y - 44 = 23.5(x - 4)\).
  • Simplifying this equation is easy: Breaking down the calculation first by expanding, \(y - 44 = 23.5x - 94\) and solving for \(y\) gives us \(y = 23.5x - 50\).
The equation \(y = 23.5x - 50\) represents the tangent line at the specified point on the curve. This algebraic expression models the function's behavior precisely at the point of tangency and can be easily graphed to visualize the tangent line against the original function.

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Most popular questions from this chapter

Let \(p, q,\) and \(r\) be functions such that \(p(z)=q(r(z))\). If \(r(3)=3, q(3)=-2, r^{\prime}(3)=4,\) and \(q(3)=6,\) find \(p(3)\) and \(p^{\prime}(3)\).

Find the first derivative. $$ F(x)=\sec 5 x \tan 5 x \sin 5 x $$

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Assuming that the equation determines a differentiable function \(f\) such that \(y=f(x),\) find \(y^{\prime}\). $$ \frac{\sqrt{x}+1}{\sqrt{y}+1}=y $$

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