jmc

We see that 1 is in the range of $f(x)=x^2-5x+c$ if and only if the equation $x^2-5x+c=1$ has a real root. We can re-write this equation as $x^2-5x+(c-1)=0$. The discriminant of this quadratic is $(-5{)}^2-4(c-1)=29-4c$. The quadratic has a real root if and only if the discriminant is nonnegative, so $29-4c\ge 0$. Then $c\le 29/4$, so the largest possible value of $c$ is $\boxed{\frac{29}{4}}$.