Question

$$\begin{array}{l}{\text { In Exercises } 9 \text { and } 10, \text { the interval } a \leq x \leq b \text { is given. Let } A=} \\ {(a, f(a)) \text { and } B=(b, f(b)) . \text { Write an equation for }} \\ {\quad(\text { a) the secant line } A B .} \\ {\text { (b) a tangent line to } f \text { in the interval }(a, b) \text { that is parallel to } A B \text { . }}\end{array}$$ $$f(x)=x+\frac{1}{x}, \quad 0.5 \leq x \leq 2$$

Short answer

The equation for the secant line AB is \(y = 2.5\). The equation of the tangent line to the function \(f\) in the interval \(0.5 \leq x \leq 2\) that is parallel to AB is \(y = 2\).

Step-by-step solution

Calculate the Coordinates of Points A and B

The end points of the interval, \(0.5\) and \(2\), correspond to the \(x\)-values for points A and B. Thus, to find the coordinates of A and B, substitute these values into the equation for \(f(x)\). For Point A, \(f(0.5) = 0.5 + \frac{1}{0.5} = 2.5\). So A = (0.5,2.5). For Point B, \(f(2) = 2 + \frac{1}{2} = 2.5\), so B = (2,2.5)

Calculate the Slope of the Secant Line AB

The slope of a line passing through two points \( (x_{1},y_{1}) \) and \( (x_{2},y_{2}) \) is given by the formula \( m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \). Using A(0.5, 2.5) and B(2, 2.5), the slope is calculated to be \( m = \frac{2.5 - 2.5}{2 - 0.5} = 0 \).

Write an equation for the Secant Line AB

The general form for the equation of a line is \( y = mx + c \), where \(m\) is the slope and \(c\) is the y-intercept. The slope of the secant is 0 and it passes through the point A(0.5, 2.5), so the equation is \( y=0*x + 2.5 = 2.5. \)

Find the Slope of \(f\) at any point in the interval \(0.5 \leq x \leq 2\)

A line is parallel to another if their slopes are equal. So we need to find a tangent to \(f\) with slope 0. The derivative of \(f\) at any point gives the slope of the tangent at that point. \(f'(x) = 1 - \frac{1}{x^2}\). For \(f'(x)\) to be 0, we need \(1 - \frac{1}{x^2} = 0 => 1 = \frac{1}{x^2} => x^2 = 1\). From this, \(x = 1\) and \(x = -1\) but only \(x = 1\) lies in the interval. So the tangent line parallel to AB touches \(f\) at (1, f(1)) = (1, 2).

Write an equation for the tangent line

The slope of this line is also 0 and it passes through the point (1,2). So \(y = 0*x + 2 = 2.\)