Question

Find equations of all tangent lines to the graph of \(f(x)=\arccos x\) that have slope -2

Short answer

The equations of the tangent lines to the graph \(f(x)=\arccos x\) that have a slope of -2 are \(y = -x + \pi -1\) and \(y = -2x + \pi +1\).

Step-by-step solution

Find the derivative

The derivative of \(f(x)=\arccos x\) can be found using the chain rule. In general, the derivative of \(\arccos u\) equals -1 divided by the square root of 1 minus \(u^2\), times \(u'\). Therefore, the derivative of the function \(f(x)\) is \(f'(x)=-\frac{1}{\sqrt{1-x^2}}\).

Substitute the slope

Since the slope of the tangent line is given as -2, we can write the equation for the derivative as \(f'(x) = -2\). Therefore, we get the equation \(-\frac{1}{\sqrt{1-x^2}} = -2\). Solving for \(x\), we get \(x=\pm \frac{1}{2}\). These are the x-coordinates at which the tangent to the function has a slope of -2.

Find the corresponding y-coordinates

The y-coordinates can be found by substituting the x-coordinates found in the previous step into the original function. So, \(f(\frac{1}{2})=\arccos(\frac{1}{2})=\frac{\pi}{3}\) and \(f(-\frac{1}{2})=\arccos(-\frac{1}{2})=\frac{2\pi}{3}\). Therefore, the points are \((\frac{1}{2}, \frac{\pi}{3})\) and \((-\frac{1}{2}, \frac{2\pi}{3})\)

Find the equations of the tangent lines

Now let's find the equation of the line using the point-slope form which is \(y-y_1 = m(x-x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line. The equation of the tangent line at the point \((\frac{1}{2}, \frac{\pi}{3})\) is \(y-\frac{\pi}{3} = -2(x-\frac{1}{2})\). After simplifying, we get the equation \(y = -2x + \pi -1\). For the point \((-\frac{1}{2}, \frac{2\pi}{3})\), the equation of the tangent line is \(y-\frac{2\pi}{3} = -2(x+\frac{1}{2})\). After simplifying, we get the equation \(y = -2x + \pi +1\).