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In Exercises \(11-18,\) find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$ f(x)=x^{2}+1, \quad(2,5) $$

Short Answer

Expert verified
The slope at the point is 4, with a tangent line equation of \( y = 4x - 3 \).

Step by step solution

01

Find the derivative

First, we need to find the derivative of the function \( f(x) = x^2 + 1 \). The derivative, denoted as \( f'(x) \), represents the slope of the function at any point \( x \). For \( f(x) = x^2 + 1 \), the derivative is obtained by applying the power rule. The derivative of \( x^2 \) is \( 2x \), and the derivative of a constant \( 1 \) is \( 0 \). Therefore, \( f'(x) = 2x \).
02

Calculate the slope at the given point

Now, we find the slope of the function at the point \((2, 5)\). To do this, substitute \( x = 2 \) into the derivative \( f'(x) = 2x \). So, \( f'(2) = 2 \times 2 = 4 \). The slope of the tangent line at \( x = 2 \) is \( 4 \).
03

Write the equation of the tangent line

We use the point-slope form of a line to write the equation of the tangent. The point-slope form 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. Here, \( m = 4 \) and \( (x_1, y_1) = (2, 5) \). Substituting these values, we get: \( y - 5 = 4(x - 2) \). Simplifying, \( y = 4x - 8 + 5 = 4x - 3 \). Therefore, the equation of the tangent line is \( y = 4x - 3 \).

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Key Concepts

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

Tangent Line
A tangent line is a straight line that just 'touches' a curve at a particular point, without crossing it. Think of it like when a pen gently grazes the surface of a page, only meeting the page at one precise point. This line is crucial because it gives us a way to approximate the behavior of functions near a given point.
The formula for finding the equation of a tangent line involves the slope and a specific point:
  • The slope of the tangent line comes from the derivative of the function evaluated at the given point.
  • The point on the tangent line is exactly the point on the curve you're interested in.
In our example, the function is given by \( f(x) = x^2 + 1 \), and we want to find the tangent line at \( (2, 5) \). We first find the derivative to determine the slope, and then use the point-slope form of a line to write the equation of the tangent line, \( y = 4x - 3 \). This line will closely approximate \( f(x) \) near \( x = 2 \).
Slope of a Function
The slope of a function at a specific point describes how quickly the function's value is changing at that point. It is a measure of steepness, much like how you might describe the incline of a hill you are walking up.
  • Positive slope: The function is increasing as you move right.
  • Negative slope: The function is decreasing as you move right.
  • Zero slope: The function is flat at that point (no change as you move right).
In mathematical terms, at any point \( x \) on a function \( f(x) \), the slope is given by the derivative \( f'(x) \). For our specific function \( f(x) = x^2 + 1 \), the derivative is \( f'(x) = 2x \). This tells us that the rate of change of \( f(x) \) is twice the \( x \)-value. At \( x = 2 \), the slope is \( 4 \), indicating the function is increasing quite steeply at that point.
Differentiation
Differentiation is a fundamental concept in calculus, primarily used to find the derivative of a function. It tells us how a function is changing at any given point, capturing its rate of change or its slope.
To differentiate a function like \( f(x) = x^2 + 1 \), we apply rules of differentiation:
  • Power Rule: For \( x^n \), the derivative is \( nx^{n-1} \).
  • Constant Rule: The derivative of a constant is \( 0 \).
Using these rules, the derivative of \( x^2 \) is \( 2x \). The constant \( 1 \) has a derivative of \( 0 \). So, the complete derivative is \( f'(x) = 2x \).
This derivative function, \( f'(x) \), gives us the slope of \( f(x) \) for any value of \( x \). Differentiation allows us to precisely understand how our original function behaves and varies, providing the groundwork for much of calculus.

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