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Find the largest real number $c$ such that $1$ is in the range of $f(x)=x^2-5x+c-3x+8$.

 Sep 6, 2023
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To find the largest real number \( c \) such that \( 1 \) is in the range of \( f(x) = x^2 - 5x + c - 3x + 8 \), we first simplify the function to: \[ f(x) = x^2 - 8x + (c + 8) \] For \( 1 \) to be in the range of \( f(x) \), there must be some value(s) of \( x \) for which \( f(x) = 1 \). This can be written as: \[ x^2 - 8x + (c + 7) = 0 \] For this equation to have at least one real root, the discriminant of the quadratic equation must be non-negative: \[ \Delta = b^2 - 4ac \] \[ \Delta = (-8)^2 - 4 * 1 * (c + 7) \] \[ \Delta = 64 - 4c - 28 \] \[ \Delta = 36 - 4c \] For \( \Delta \) to be non-negative, \( 36 - 4c \geq 0 \), or \( c \leq 9 \). Therefore, the largest real number \( c \) for which \( 1 \) is in the range of \( f(x) \) is c = 9.

 Sep 6, 2023

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