## Calculus: Early Transcendentals 8th Edition

$f(4)=3$ and $f'(4)=\frac{1}{4}$
The tangent line $(l):y=f(x)$ is at point $A(4,3)$, so point $A(4,3)$ also lies in $(l)$. Therefore, $f(4)=3$ The equation of the tangent line $l$ would have the following form: $$(l): y=ax+b$$ Since $l$ passes through point $A(4,3)$, we apply the equation of $l$ to $A$, which means $$4a+b=3\hspace{1cm}(1)$$ $l$ also passes through point $B(0,2)$, we also can apply the equation of $l$ to $B$, which means $$0a+b=2$$$$b=2\hspace{1cm}(2)$$ Apply (2) to (1), we have $$4a+2=3$$$$a=\frac{1}{4}$$ Since $a$ is the slope of the tangent line $l$ at point $A(4,3)$, $f'(4)=a=\frac{1}{4}$