Find the largest real number $c$ such that $1$ is in the range of $f(x)=x^2-5x+c+x^2-8x$.
We have $f(x)=2x^2-13x+c$. The range of $f$ is $\{y \in \mathbb{R}\mid y \ge d\}$, where $d$ is the discriminant of the quadratic.
Because the quadratic has a positive leading coefficient, the range is $[d,\infty).$ The discriminant is $(-13)^2-4(2)(c)=169-8c$, and for $d \ge 1$ we must have $169-8c \le 0.$ Solving gives $c \ge 21.125,$ so the largest such $c$ is $\boxed{21.125}$.
To find the largest real number $c$ such that $1$ is in the range of $f(x)=x^2-5x+c+x^2-8x$, we need to determine the values of $c$ for which the quadratic equation $f(x) = 1$ has real solutions.
First, let's rewrite the given quadratic equation:
$$f(x) = x^2-5x+c+x^2-8x.$$
Combining like terms, we can simplify the equation to:
$$f(x) = 2x^2-13x+c.$$
To find the values of $x$ that satisfy $f(x) = 1$, we set up the following equation:
$$2x^2-13x+c = 1.$$
Now we have a quadratic equation. To find the values of $x$, we can rearrange the equation to be in the form of $ax^2+bx+c=0$:
$$2x^2-13x+(c-1) = 0.$$
For this quadratic equation to have real solutions, the discriminant ($b^2-4ac$) must be non-negative. The discriminant is given by:
$$D = (-13)^2 - 4(2)(c-1).$$
To determine the largest value of $c$, we want the discriminant to be equal to zero:
$$D = 0.$$
Setting $D$ equal to zero, we get:
$$(-13)^2 - 4(2)(c-1) = 0.$$
Simplifying this equation, we have:
$$169 - 8(c-1) = 0.$$
Expanding and further simplifying, we get:
$$169 - 8c + 8 = 0.$$
Combining like terms, we have:
$$8 - 8c = 0.$$
Solving for $c$, we get:
$$8c = 8.$$
$$c = 1.$$
Therefore, the largest real number $c$ for which $1$ is in the range of $f(x)$ is $c = 1$.