Problem 18
Question
Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$f(x)=\sqrt{x+1}, \quad\quad(8,3)$$
Step-by-Step Solution
Verified Answer
The slope is \(\frac{1}{6}\) and the tangent line is \(y = \frac{1}{6}x + \frac{5}{3}\).
1Step 1: Find the derivative of the function
To find the slope of the tangent line, you first need the derivative of the function. The function is given by \( f(x) = \sqrt{x+1} \), which can be rewritten as \( f(x) = (x+1)^{1/2} \).Using the power rule, the derivative is \( f'(x) = \frac{1}{2} (x+1)^{-1/2} \).
2Step 2: Evaluate the derivative at the given point
Now, substitute \(x = 8\) into the derivative to find the slope of the tangent line at the point (8, 3).\[ f'(8) = \frac{1}{2} (8+1)^{-1/2} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \]Hence, the slope of the tangent line at \((8, 3)\) is \( \frac{1}{6} \).
3Step 3: Use the point-slope form to find the equation of the tangent line
The point-slope form of the line equation is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point. Substituting in the values, we have:\[ y - 3 = \frac{1}{6}(x - 8) \]Simplify it to get the equation of the tangent line:\[ y = \frac{1}{6}x - \frac{4}{3} + 3 \]\[ y = \frac{1}{6}x + \frac{5}{3} \].
Key Concepts
derivativeslope of tangent linetangent line equation
derivative
The concept of a derivative is central to calculus. It represents the rate of change of a function with respect to a variable. To put it simply, the derivative gives us the slope of the tangent line to the graph of a function at any given point. This tells us how steep the graph is at that exact location.
In mathematical terms, if you have a function \( f(x) \), the derivative, denoted by \( f'(x) \), describes how \( f(x) \) changes as \( x \) changes. In the provided exercise, the function \( f(x) = \sqrt{x+1} \) is rewritten using ein power form: \( f(x) = (x+1)^{1/2} \). Using the power rule from calculus, we find the derivative: \( f'(x) = \frac{1}{2} (x+1)^{-1/2} \). This tells us the rate at which \( \sqrt{x+1} \) changes as \( x \) increases.
Understanding derivatives is essential for analyzing how functions behave, and they serve as the foundation for many calculus applications in physics, engineering, economics, and beyond.
In mathematical terms, if you have a function \( f(x) \), the derivative, denoted by \( f'(x) \), describes how \( f(x) \) changes as \( x \) changes. In the provided exercise, the function \( f(x) = \sqrt{x+1} \) is rewritten using ein power form: \( f(x) = (x+1)^{1/2} \). Using the power rule from calculus, we find the derivative: \( f'(x) = \frac{1}{2} (x+1)^{-1/2} \). This tells us the rate at which \( \sqrt{x+1} \) changes as \( x \) increases.
Understanding derivatives is essential for analyzing how functions behave, and they serve as the foundation for many calculus applications in physics, engineering, economics, and beyond.
slope of tangent line
Once you have the derivative of a function, you can find the slope of the tangent line at any specific point. The slope of the tangent line represents the exact incline of the curve at that point, letting you know the immediate direction in which the function is heading.
In our case, the derivative of \( f(x) = \sqrt{x+1} \) is \( f'(x) = \frac{1}{2} (x+1)^{-1/2} \). To find the slope of the tangent at the point \((8,3)\), we simply substitute \( x = 8 \) into the derivative. The calculation \( f'(8) = \frac{1}{2} (8+1)^{-1/2} = \frac{1}{6} \) shows that the slope of the tangent line at this point is \( \frac{1}{6} \).
Knowing the slope helps you understand the behavior of the function locally around the point of interest. It's like zooming into a curve at that point and seeing a straight path that best represents the function right there.
In our case, the derivative of \( f(x) = \sqrt{x+1} \) is \( f'(x) = \frac{1}{2} (x+1)^{-1/2} \). To find the slope of the tangent at the point \((8,3)\), we simply substitute \( x = 8 \) into the derivative. The calculation \( f'(8) = \frac{1}{2} (8+1)^{-1/2} = \frac{1}{6} \) shows that the slope of the tangent line at this point is \( \frac{1}{6} \).
Knowing the slope helps you understand the behavior of the function locally around the point of interest. It's like zooming into a curve at that point and seeing a straight path that best represents the function right there.
tangent line equation
The equation of a tangent line is crucial because it provides the linear approximation of the function at a specific point. This approximation is incredibly useful in various mathematical, scientific, and engineering contexts.
To find the equation of the tangent line, we use the point-slope form: \( y - y_1 = m(x - x_1) \). Here, \( m \) is the slope of the tangent found earlier, and \((x_1, y_1)\) is the given point, which in this case is \((8, 3)\).
Plugging in our values, we have \( y - 3 = \frac{1}{6}(x - 8) \). By simplifying, we get the tangent line equation \( y = \frac{1}{6}x + \frac{5}{3} \).
This linear equation describes a straight line that just touches the curve of \( f(x) = \sqrt{x+1} \) at \( (8, 3) \), and it matches the function's slope precisely at that point. Understanding how to derive this equation not only gives insights into the function's behavior at one point but also forms the basis for more complex analyses in calculus.
To find the equation of the tangent line, we use the point-slope form: \( y - y_1 = m(x - x_1) \). Here, \( m \) is the slope of the tangent found earlier, and \((x_1, y_1)\) is the given point, which in this case is \((8, 3)\).
Plugging in our values, we have \( y - 3 = \frac{1}{6}(x - 8) \). By simplifying, we get the tangent line equation \( y = \frac{1}{6}x + \frac{5}{3} \).
This linear equation describes a straight line that just touches the curve of \( f(x) = \sqrt{x+1} \) at \( (8, 3) \), and it matches the function's slope precisely at that point. Understanding how to derive this equation not only gives insights into the function's behavior at one point but also forms the basis for more complex analyses in calculus.
Other exercises in this chapter
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