Positive and Negative Domains: Comparisons to zero or between parabolas

To solve the problem of finding when $f(x) = -x^2 + 5x + 6 > 0$, follow these steps:

Step 1: Find the roots of the quadratic equation $-x^2 + 5x + 6 = 0$. Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -1$, $b = 5$, and $c = 6$.

Step 2: Calculate the discriminant:

$b^2 - 4ac = 5^2 - 4(-1)(6) = 25 + 24 = 49$

Step 3: Solve for $x$:

$x = \frac{-5 \pm \sqrt{49}}{-2}$

Step 4: Simplify the roots:

$x = \frac{-5 \pm 7}{-2}$

This gives the solutions:

$x = \frac{-5 + 7}{-2} = -1 \quad \text{and} \quad x = \frac{-5 - 7}{-2} = 6$

Step 5: Evaluate intervals defined by these roots: $(- \infty, -1)$, $(-1, 6)$, and $(6, \infty)$.

Step 6: Test a sample point in each interval to check where $f(x) > 0$:

• For interval $(- \infty, -1)$, choose $x = -2$. $f(-2) = -(-2)^2 + 5(-2) + 6 = -4 - 10 + 6 = -8$ (negative)
• For interval $(-1, 6)$, choose $x = 0$. $f(0) = -(0)^2 + 5(0) + 6 = 6$ (positive)
• For interval $(6, \infty)$, choose $x = 7$. $f(7) = -(7)^2 + 5(7) + 6 = -49 + 35 + 6 = -8$ (negative)

Step 7: Conclude that the function is positive in the interval: $-1 < x < 6$.

Therefore, the solution is $-1 < x < 6$.

To determine where $f(x) = x^2 - 6x + 8$ is less than zero, we need to find the roots of the quadratic equation and test the intervals determined by them.

Step 1: Factor the quadratic.
The equation can be rewritten as $x^2 - 6x + 8 = (x - 2)(x - 4) = 0$.
Thus, the roots are $x = 2$ and $x = 4$.

Step 2: Using these roots, we can identify intervals to test where the product $(x - 2)(x - 4) < 0$. The intervals derived from the roots are:

• Interval 1: $x < 2$
• Interval 2: $2 < x < 4$
• Interval 3: $x > 4$

Step 3: Test each interval to find where $f(x) < 0$.
- For $x < 2$, both factors $(x - 2)$ and $(x - 4)$ are negative, thus their product is positive.
- For $2 < x < 4$, $(x - 2)$ is positive, and $(x - 4)$ is negative, so their product is negative.
- For $x > 4$, both $(x - 2)$ and $(x - 4)$ are positive, so their product is positive.

Therefore, the function $f(x) = x^2 - 6x + 8$ is less than zero for the range $2 < x < 4$.

Thus, the values for which $f(x) < 0$ is true are $2 < x < 4$.

The given function is $f(x) = -2x^2 + 4x - 6$.

First, we find the roots of the equation $f(x) = 0$, which are the critical points where the function can change its sign.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -2$, $b = 4$, and $c = -6$:

Calculate the discriminant $\Delta = b^2 - 4ac = 4^2 - 4(-2)(-6) = 16 - 48 = -32$.

The discriminant is negative, indicating there are no real roots.

Because the parabola opens downwards and there are no x-intercepts (real roots), the function does not cross the x-axis.

Hence, for all real numbers, $f(x) \leq 0$.

Therefore, there are no values of $x$ for which $f(x) > 0$.

Thus, the solution is: No answer.