Solve f(x)=x²-6x+8 > -x+4: Quadratic-Linear Function Comparison

The following function is graphed below:

$f(x)=x^2-6x+8$

$g(x)=-x+4$

For which values of x is
$f(x) > g(x)$ true?

To determine for which values of $x$ the condition $f(x) > g(x)$ holds, follow these steps:

• Step 1: Set up the inequality $x^2 - 6x + 8 > -x + 4$.
• Step 2: Rearrange all terms to one side: $x^2 - 6x + 8 + x - 4 > 0$ simplifies to $x^2 - 5x + 4 > 0$.
• Step 3: Solve the related quadratic equation $x^2 - 5x + 4 = 0$. This factors as $(x - 1)(x - 4) = 0$, giving roots $x = 1$ and $x = 4$.
• Step 4: Determine intervals defined by the roots: $(-\infty, 1)$, $(1, 4)$, and $(4, \infty)$.
• Step 5: Test points from each interval in the inequality $x^2 - 5x + 4 > 0$:
  • For $x = 0$ (from interval $(-\infty, 1)$), $x^2 - 5x + 4 = 4$ which is greater than 0.
  • For $x = 2$ (from interval $(1, 4)$), $x^2 - 5x + 4 = -2$ which is less than 0.
  • For $x = 5$ (from interval $(4, \infty)$), $x^2 - 5x + 4 = 9$ which is greater than 0.
• Step 6: From the test points, determine $f(x) > g(x)$ in intervals $(-\infty, 1)$ and $(4, \infty)$.

Thus, $f(x) > g(x)$ for $x < 1$ or $x > 4$.

Therefore, the solution to the problem is $x < 1, 4 < x$, corresponding to choice 2.