Question

Finding an Equation of a Tangent Line In Exercises \(33-38,\) find an equation of the line that is tangent to the graph of \(f\) and parallel to the given line. $$\begin{array}{ll}{\text { Function }} & {\text { Line }} \\\ {f(x)=\frac{1}{\sqrt{x-1}}} & {x+2 y+7=0}\end{array}$$

Short answer

The equation of the line that is tangent to the function \(f(x)=\frac{1}{\sqrt{x-1}}\) and parallel to the line \(x + 2y + 7 = 0\) is \(y = - \frac{1}{2}x + 2\).

Step-by-step solution

Find slope of the line

The given line is in the form \(x + 2y + 7 = 0\). Rearranging it to the slope-intercept form \(y = mx + c\), we get \(y = - \frac{1}{2}x - \frac{7}{2}\). The slope of this line is \(- \frac{1}{2}.\ Since parallel lines have equal slopes, the slope of the tangent line is also - \frac{1}{2}.

Differentiate the Function

Take the derivative of the function \(f(x) = \frac{1}{\sqrt{x - 1}} = (x-1)^{-1/2}\), using the power rule for differentiation. This yields \(f'(x) = -1/2 * (x - 1)^{-3/2}\). This expression represent the slope of the tangent line at a certain point \(x\). Since we want this to be equal to \(-1/2\), we need to solve the equation \(f'(x) = -1/2\).

Solve for x

By setting \(f'(x) = -1/2\), we get \(-1/2 = -1/2 * (x - 1)^{-3/2}\). By cross-multiplication, we obtain \((x - 1)^{-3/2} = 1 \Rightarrow (x - 1) = 1\). This gives \(x = 2\).

Find the y-coordinate

Plug this \(x\)-value into the function \(f(x)\) to find the corresponding \(y\)-value. This gives \(f(2) = \frac{1}{\sqrt{2 - 1}} = 1.\ So, the line touches the graph at the point (2,1).

Find the equation of the tangent line

Use the point-slope form of a line, interested in the point (2,1) and slope -1/2. This yields \(y - 1 = - \frac{1}{2} (x - 2)\), which simplifies to \(y = - \frac{1}{2}x + 2\).

Key concepts

Slope of a Line

The slope of a line is a measure of how steep it is. It's calculated as the rise over the run, or the change in y over the change in x. In mathematical terms, if two points on the line are \( (x_1, y_1) \) and \( (x_2, y_2) \), the slope m is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \).

This concept is foundational when finding the equation of a tangent line, because the slope of the tangent line at any point is equivalent to the slope of the curve at that point. When a problem involves a line that is parallel to the tangent, as in our exercise, the slopes of both lines will be the same, allowing us to set this common slope value as our target in finding the tangent line equation.

Derivative of a Function

The derivative of a function is a fundamental tool in calculus that measures how a function's output value changes as its input value changes. Concepts like velocity and acceleration are practical examples of derivatives in the real world. Mathematically, if you have a function \( f(x) \), the derivative of this function is denoted \( f'(x) \) or \( \frac{df}{dx} \), and it gives us the slope of the tangent line to the curve at any point x.

In the context of our exercise, after finding the derivative \( f'(x) \), we set it equal to the slope we found for the given line. This equation helps us determine at which point x the slope of the function's curve matches the slope of the given line, indicating where the tangent line will touch the curve.

Power Rule for Differentiation

The power rule is a quick and handy technique for differentiating functions of the form \( x^n \), where n is any real number. The rule states that the derivative of \( x^n \) with respect to x is \( nx^{n-1} \).

To apply the power rule as we saw in Step 2 of our exercise, we rewrite the function \( f(x) = \frac{1}{\sqrt{x-1}} \) as \( f(x) = (x-1)^{-\frac{1}{2}} \) and then differentiate using the power rule to get \( f'(x) = -\frac{1}{2}(x-1)^{-\frac{3}{2}} \). This expresses the slope of the tangent line in terms of x, allowing us to solve for the exact point where the function's slope matches that of the given line.

Slope-Intercept Form

The slope-intercept form is an equation of a line written as \( y = mx + b \), where m represents the slope, and b represents the y-intercept—the point where the line crosses the y-axis. This form is particularly useful for graphing linear equations and understanding the relationship between the slope and y-intercept.

When solving problems like our exercise, converting the given line into slope-intercept form allows us to easily identify the slope. Then, by plugging the coordinates of the point where the tangent meets the curve and the slope into the slope-intercept equation, we can solve for b and obtain the full equation of the tangent line.