Question

Find the tangent line approximation for sqrt7+x near x=2.

Answer 1

Let \(f(x)=\sqrt{7+x}\). At \(x=2\), the slope of the tangent line is
\[f'(2)=\frac{1}{2\sqrt{7+2}}=\frac{1}{6}\]
At \(x=2\), the function value is \(f(2)=\sqrt{7+2}=\sqrt9=3\).

Hence the tangent line is given by
\[y=3+\frac{1}{6}(x-2)=\frac{1}{6}x+\frac{8}{3}\]