Question

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{y=4-x^{2}-\ln \left(\frac{1}{2} x+1\right)} \quad \frac{\text { Point }}{\left(0,4\right)}\)

Short answer

The equation of the tangent line to the graph of the function at the point (0,4) is \(y=-0.5x+4\)

Step-by-step solution

Find the derivative

First, find the derivative of the function \(y=4-x^{2}-\ln \left(\frac{1}{2} x+1\right)\) using the power rule and the chain rule. The derivative of \(4-x^{2}\) is \(-2x\) and the derivative of \(-\ln (\frac{1}{2}x+1)\) is \(-\frac{1}{2x+2}\). Adding up all parts the derivative \(f'(x)\) becomes \(f'(x)=-2x-\frac{1}{2x+2}\)

Find the slope of the tangent line

Using the previously found derivative, find the slope of the tangent line at the given point (0,4), by replacing the x-coordinate into the derivative. Hence, \(f'(0)=0-\frac{1}{2(0)+2}= -0.5\)

Find the equation of the tangent line

The equation of the tangent line can be found using the point-slope form \(y-y_1=m(x-x_1)\), where \(m\) is the slope and \(x_1,y_1\) are the x and y coordinates of the given point (0,4), respectively. By substituting these values into the equation, it becomes \(y-4=-0.5(x-0)\), which simplifies to \(y=-0.5x+4\)

Confirm results using a graphing utility

Plot the curve of original function \(y=4-x^2-\ln(0.5x+1)\) and its tangent line \(y=-0.5x+4\) at point (0,4). It can be confirmed that the tangent line obtained is correct if it only touches the function at the given point (0,4). Additionally, with the use of the derivative feature of a graphing utility, one can confirm the obtained slope of the tangent line.