Question

Find an equation of the line that is tangent to the graph of \(f\) and parallel to the given line. Function \(\quad\) Line \(f(x)=\frac{1}{\sqrt{x-1}} \quad x+2 y+7=0\)

Short answer

The equation of the line that is tangent to the graph of the given function and parallel to the given line is \(x+2y-4=0\).

Step-by-step solution

Find the Derivative of the Function

We use the power rule for differentiation, which is (u^n)'=n*u^(n-1)*u'. Here, \(u=x-1\) and \(n=-0.5\). So, the derivative \(f'(x)\) is computed as: \[\frac{d}{dx}\left(\frac{1}{\sqrt{x-1}}\right)=-\frac{1}{2}(x-1)^{-1.5}\]

Find the Slope of the Given Line

The given equation of the line is \(x+2y+7=0\). In the slope-intercept form \(y=mx+c\), m is the slope. You rearrange the given equation to this form: \[y=-\frac{1}{2}x-\frac{7}{2}\]. It gives the slope is \(-\frac{1}{2}\).

Find the x-coordinate of the Contact Point

Because the tangent line is parallel to the given line, it has the same slope and at the point of tangency, the derivative of the function will be equal to the slope of the tangent line. Therefore, we set \(f'(x)\) from step 1 to \(-\frac{1}{2}\) from step 2, and solve for \(x\): \[-\frac{1}{2}(x-1)^{-1.5}=-\frac{1}{2}\]. This gives \(x=2\].

Find the Equation of the Tangent Line

We've the slope from step 2 as \(-\frac{1}{2}\), and the point of contact which is the x-value in step 3 and the y-value can be found by plugging the x-value into the original function \(f(x)\), which gives \((2,1)\). The equation of the line is then given by the point-slope form: \(y-y_1=m(x-x_1)\), where \(m\) is the slope and \((x_1,y_1)\) is the point of contact. Plug the slope and the point of contact values into this formula, that gives the equation of the tangent line as \(y-1=-\frac{1}{2}(x-2)\). Rearranging this to standard form gives the equation as \(x+2y-4=0\).