Answer
$y=-\displaystyle \frac{1}{2}x+4$
Work Step by Step
Use:
Theorem 5.2 :properties of ln(x),
Theorem 5.3:$\ \ \ (\displaystyle \ln x)^{\prime}=\frac{1}{x}, \ \ \ (\displaystyle \ln u)=\frac{u^{\prime}}{u}$
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$y=f(x)=4-x^{2}-\displaystyle \ln(\frac{1}{2}x+1)$
$f^{\prime}(x)=0-2x-\displaystyle \frac{\frac{1}{2}}{(\frac{1}{2}x+1)}=-2x-\frac{1}{x+2}$
The slope of the tangent at ($0,4$) is $f^{\prime}(0)=0-\displaystyle \frac{1}{2}=-\frac{1}{2}$
Point-slope equation of the tangent line:
$y-y_{0}=m(x-x_{0})$
$y-4=-\displaystyle \frac{1}{2}(x-0)$
$y=-\displaystyle \frac{1}{2}x+4$
For (b) and (c), see image attached.
