Chapter 3: Problem 50
a. For the following functions, find \(f^{\prime}\) using the definition. b. Determine an equation of the line tangent to the graph of \(f\) at \((a, f(a))\) for the given value of \(a\) $$f(x)=\sqrt{x+2} ; a=7$$
Short Answer
Expert verified
Answer: The equation of the tangent line is \(y = \frac{1}{6}x + \frac{11}{6}\).
Step by step solution
01
Definition of the derivative
Let's recall the definition of the derivative of a function, \(f^{\prime}(x)\). It is given by the following limit: $$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$Now, we'll apply this definition to the given function, \(f(x) = \sqrt{x + 2}\).
02
Plug in the function into the definition
Plugging the function into the definition, we get the following:
$$f^{\prime}(x) = \lim_{h \to 0} \frac{\sqrt{(x + h) + 2} - \sqrt{x + 2}}{h}$$
03
Rationalize the numerator
To solve the limit, we can rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator:
$$f^{\prime}(x) = \lim_{h \to 0} \frac{\sqrt{(x + h) + 2} - \sqrt{x + 2}}{h} \cdot \frac{\sqrt{(x + h) + 2} + \sqrt{x + 2}}{\sqrt{(x + h) + 2} + \sqrt{x + 2}}$$
Multiplying the numerators, we'll obtain a difference of squares:
$$f^{\prime}(x) = \lim_{h \to 0} \frac{(x + h) + 2 - (x + 2)}{h (\sqrt{(x + h) + 2} + \sqrt{x + 2})}$$
04
Simplify and solve the limit
Now, we can simplify the numerator and cancel out the common factor in the numerator and the denominator:
$$f^{\prime}(x) = \lim_{h \to 0} \frac{h}{h (\sqrt{(x + h) + 2} + \sqrt{x + 2})}$$
$$f^{\prime}(x) = \lim_{h \to 0} \frac{1}{\sqrt{(x + h) + 2} + \sqrt{x + 2}}$$
As \(h \to 0\), we get:
$$f^{\prime}(x) = \frac{1}{\sqrt{(x + 0) + 2} + \sqrt{x + 2}}$$
So, the derivative of \(f(x)\) is:
$$f^{\prime}(x) = \frac{1}{2\sqrt{x + 2}}$$
05
Determine the tangent line
Now that we have the derivative, we can find the slope of the tangent line at \(a = 7\). Using the derivative, we find the slope of the tangent line \(m\):
$$m = f^{\prime}(a) = \frac{1}{2\sqrt{7 + 2}} = \frac{1}{6}$$
To find the equation of the tangent line, we also need the coordinates of the point \((a, f(a))\). We have:
$$f(7) = \sqrt{7 + 2} = 3$$
So, the point is \((7, 3)\). Now, we will use the point-slope form of the line with point \((7, 3)\) and slope \(m = \frac{1}{6}\):
$$y - y_1 = m (x - x_1)$$
Plugging in our values:
$$y - 3 = \frac{1}{6} (x - 7)$$
And finally, we can distribute and rearrange the equation:
$$y = \frac{1}{6}x - \frac{7}{6} + 3$$
So, the equation of the tangent line is:
$$y = \frac{1}{6}x + \frac{11}{6}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Definition of Derivative
Understanding the concept of derivatives is fundamental to calculus, and it all starts with the limit definition. In essence, a derivative represents how a function's output changes as its input changes, which is practically the slope of the function at any given point.
The formal definition is as follows: the derivative of a function at a certain point is the limit of the average rate of change of the function over an interval as the interval shrinks to zero. Mathematically, it is expressed as \[f^{\textprime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\] The 'h' represents a tiny interval we add to 'x', and as it approaches zero, we capture the instantaneous rate of change of the function at point 'x'. This concept is the bedrock of finding the behavior of functions and allows us to explore changes in various scientific fields, from physics to economics.
The formal definition is as follows: the derivative of a function at a certain point is the limit of the average rate of change of the function over an interval as the interval shrinks to zero. Mathematically, it is expressed as \[f^{\textprime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\] The 'h' represents a tiny interval we add to 'x', and as it approaches zero, we capture the instantaneous rate of change of the function at point 'x'. This concept is the bedrock of finding the behavior of functions and allows us to explore changes in various scientific fields, from physics to economics.
Tangent Line Equation
A tangent line to a curve at a given point is the straight line that just 'touches' the curve at that point. This line is a good linear approximation of the function near that point. Its slope, given by the derivative at that particular point, tells us how steeply the curve changes direction.
The tangent line's equation at a point \(a, f(a)\) can be constructed using the point-slope form: \[y - f(a) = f^{\textprime}(a)(x - a)\] where \(f^{\textprime}(a)\) is the slope of the function at 'x = a', also known as the derivative at 'a'. To put this equation to use, you'd need the y-coordinate which is \(f(a)\), and the derivative \(f^{\textprime}(a)\). With those, you can effectively determine the equation of the line that skims the curve at just one point without crossing it.
The tangent line's equation at a point \(a, f(a)\) can be constructed using the point-slope form: \[y - f(a) = f^{\textprime}(a)(x - a)\] where \(f^{\textprime}(a)\) is the slope of the function at 'x = a', also known as the derivative at 'a'. To put this equation to use, you'd need the y-coordinate which is \(f(a)\), and the derivative \(f^{\textprime}(a)\). With those, you can effectively determine the equation of the line that skims the curve at just one point without crossing it.
Rationalizing the Numerator
Occasionally, when dealing with limits and derivatives, you'll encounter a difference between square roots in the numerator, which can be tricky. To make these easier to simplify, you can multiply the fraction by a convenient form of one that will eliminate the square root. This process is called rationalizing the numerator.
You do this by multiplying the numerator and the denominator by the conjugate of the numerator. If the numerator is \(\sqrt{a} - \sqrt{b}\), its conjugate is \(\sqrt{a} + \sqrt{b}\). The product of these is a difference of squares which simplifies to \(a-b\), essentially rationalizing the numerator. This technique often simplifies the expression and reveals cancellations that help in finding the limit for the derivative.
You do this by multiplying the numerator and the denominator by the conjugate of the numerator. If the numerator is \(\sqrt{a} - \sqrt{b}\), its conjugate is \(\sqrt{a} + \sqrt{b}\). The product of these is a difference of squares which simplifies to \(a-b\), essentially rationalizing the numerator. This technique often simplifies the expression and reveals cancellations that help in finding the limit for the derivative.
Difference of Squares Technique
The difference of squares is a technique vital to algebra, and calculus often involves algebraic manipulation. It's used to simplify expressions, especially when multiplying conjugate binomials like \(\sqrt{a} + \sqrt{b}\) and \(\sqrt{a} - \sqrt{b}\). According to this technique, the product will be \(a - b\).
In the context of calculating derivatives, this technique can help with rationalizing the numerator. Once the squares are subtracted, what often follows is the cancellation of terms with the 'h' in the denominator, which was preventing you from taking the limit as 'h' approaches zero. Mastering this technique not only aids in finding derivatives but also is useful in many areas of mathematics where simplification is required.
In the context of calculating derivatives, this technique can help with rationalizing the numerator. Once the squares are subtracted, what often follows is the cancellation of terms with the 'h' in the denominator, which was preventing you from taking the limit as 'h' approaches zero. Mastering this technique not only aids in finding derivatives but also is useful in many areas of mathematics where simplification is required.