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Chapter 1: Problem 16

Find the equation of the tangent line to \(y=x^{2}\) at the point where \(x=2.1\).

Short Answer

Expert verified
The equation of the tangent line is y = 4.2x - 4.41.

Step by step solution

01

- Find the derivative of the function

To find the equation of the tangent line, first, determine the derivative of the function. The function given is y = x^{2}. The derivative of y with respect to x is found using the power rule: y' = 2x.
02

- Evaluate the derivative at the given point

Substitute the x-value of the point where the tangent line is to be found (x=2.1) into the derivative to find the slope of the tangent line at that point: y' = 2(2.1) = 4.2.
03

- Find the y-coordinate of the point

Substitute x=2.1 into the original function to get the y-coordinate of the point: y = (2.1)^{2} = 4.41. So, the point is (2.1, 4.41).
04

- Use the point-slope form to find the equation of the tangent line

The point-slope form of the equation of a line is: y - y1 = m(x - x1) where m is the slope, and (x1, y1) is the point. From the previous steps, m=4.2 and (x1, y1) = (2.1, 4.41). Substitute these values into the equation: y - 4.41 = 4.2(x - 2.1).
05

- Simplify the equation

Distribute and simplify to put the equation in slope-intercept form (y=mx+b): y - 4.41 = 4.2x - 8.82 Add 4.41 to both sides: y = 4.2x - 4.41.

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Key Concepts

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

Derivative
Understanding derivatives is crucial in calculus, as they are used to determine the slope of a function at any point. The derivative of a function represents the rate of change or the slope of the function.

The given function is \(y = x^{2}\). To find the derivative, we use the power rule. The power rule states that for \(y=x^n\), the derivative \(y'\) is \(nx^{n-1}\). So for \(y = x^{2}\), the derivative is:

\(y' = 2x\).
Point-Slope Form
The point-slope form of a linear equation is another essential concept when dealing with tangent lines. The point-slope form is written as:

\(y - y_1 = m(x - x_1)\)

In this equation:
  • \(m\) is the slope of the line
  • \(x_1, y_1\) is a point on the line

From the example problem, we've already found that the slope \(m = 4.2\) and the point \( (2.1, 4.41)\) is on the curve. Plugging these values into the point-slope form, we get:

\(y - 4.41 = 4.2(x - 2.1)\).
Calculus
Calculus is the branch of mathematics that studies how things change. It includes concepts like derivatives and integrals. Derivatives help us understand the behavior of functions and their rates of change, while integrals can be used to find areas under curves and accumulations.

In the given problem, using calculus allowed us to find the slope of the function \(y = x^{2}\) at \(x = 2.1\), giving us the needed information to write the equation of the tangent line. This process is a powerful application of calculus in analyzing and understanding functions.

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Most popular questions from this chapter

In Exercises 37-48, use limits to compute \(f^{\prime}(x)\). [Hint: In Exercises \(45-48\), use the rationalization trick of Example \(8 .]\) \(f(x)=3 x+1\)

Compute the difference quotient $$ \frac{f(x+h)-f(x)}{h} . $$ Simplify your answer as much as possible. \(f(x)=x^{2}-7\)

Compute the following limits. \(\lim _{x \rightarrow \infty} \frac{5 x+3}{3 x-2}\)

Use limits to compute \(f^{\prime}(x)\). [Hint: In Exercises \(45-48\), use the rationalization trick of Example \(8 .]\) \(f(x)=\frac{1}{\sqrt{x}}\)

Use a derivative routine to obtain the value of the derivative. Give the value to 5 decimal places. \(f^{\prime}(0)\), where \(f(x)=10^{1+x}\)
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