/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 a. For the following functions a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

a. For the following functions and points, find \(f^{\prime}(a)\) b. Determine an equation of the line tangent to the graph of \(f\) at \((a, f(a))\) for the given value of \(a\). $$f(x)=2 x^{3} ; a=10$$

Short Answer

Expert verified
The equation of the line tangent to the graph at \(a=10\) is: \(y - 2000 = 600(x - 10)\)

Step by step solution

01

Find the derivative of the function \(f(x)\)

To determine the derivative of the function \(f(x) = 2x^3\), we will use the power rule. The power rule states that if \(f(x) = x^n\), where \(n\) is a constant, then \(f'(x) = nx^{n-1}\). Applying the power rule to our function, we get: $$f^{\prime}(x) = 3 \cdot 2x^{3-1} = 6x^2$$
02

Evaluate the derivative at \(a = 10\)

To find the slope of the tangent line at \(a = 10\), we will substitute the value of \(a\) into the derivative: $$f^{\prime}(a) = 6a^2 = 6(10)^2 = 600$$ So, the slope of the tangent line at \(a = 10\) is 600.
03

Find the point \((a, f(a))\)

Now we will find the coordinates of the point on the graph of \(f(x)\) where the tangent line touches it. Since \(a = 10\), we will evaluate \(f(a)\): $$f(a) = 2a^3 = 2(10)^3 = 2000$$ Therefore, the point \((a, f(a))\) is \((10, 2000)\).
04

Determine the equation of the tangent line

Finally, we will use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) are the coordinates of the point on the line and \(m\) is the slope. In our case, \((x_1, y_1) = (10, 2000)\) and \(m = 600\). Plugging in these values, we get: $$y - 2000 = 600(x - 10)$$ And that is the equation of the line tangent to the graph of \(f(x)=2x^3\) at \(a = 10\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Power Rule
The power rule is a fundamental concept in calculus, used to find the derivative of functions that are expressed as powers. Essentially, it makes taking derivatives simple for polynomial expressions. Whether you’re dealing with a term like \(x^3\) or any variable raised to a power, the power rule states that if you have a function \(f(x) = x^n\), its derivative \(f'(x)\) would be \(nx^{n-1}\). This involves:
  • Multiplying the power 'n' by the coefficient of the term.
  • Reducing the power by one to form the new exponent.
In our example, with the function \(f(x) = 2x^3\), applying the power rule means taking 3 (the exponent), multiplying it by 2 (the coefficient), and reducing the exponent by 1. Therefore, \(f'(x)\) becomes \(6x^2\). The power rule helps us easily compute derivatives, which are crucial in determining tangent lines and rates of change.
Tangent Line
A tangent line is a straight line that touches a curve at a single point, matching the curve's slope at that point. Understanding tangent lines is important in many fields, both mathematical and applied, from analyzing graphs to modeling real-world data. When you have a function like \(f(x) = 2x^3\), the tangent line at any point \((a, f(a))\) gives you an approximation of the curve near that point. The slope of this tangent line is given by the derivative, \(f'(a)\).
  • First, find the derivative of the function to obtain the slope.
  • Then, evaluate this derivative at the desired point to get the specific slope at that point.
  • Use this slope in conjunction with the point to determine the equation of the tangent line.
In our specific case, with \(a = 10\), the slope of the tangent line is found to be 600 through the derivative \(f'(a) = 6(10)^2 = 600\). Thus, the tangent line approximates \(f(x)\) near \((10, 2000)\). It provides a linear approach to understanding how the curve behaves around that point.
Point-Slope Form
The point-slope form is a formula used to find the equation of a line when you know the slope and a point on the line. It’s especially useful when dealing with tangent lines in calculus, as it makes finding linear equations straightforward. The formula is:\[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) are the coordinates of the point and \(m\) is the slope. To use this form:
  • First, determine the slope of the line using the derivative of the function.
  • Identify the specific point on the curve that intersects with the tangent line.
  • Substitute both the point and the slope into the point-slope formula to get the tangent line's equation.
In the case where \(a = 10\), the slope \(m\) is 600, and the point on the line is \((10, 2000)\). Thus, using the point-slope formula, you find: \[ y - 2000 = 600(x - 10) \]This calculated equation represents the tangent line, providing a clear linear relationship between \(x\) and \(y\) close to the point on the original curve.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.