To analyze the function $f(x) = 2 - 6x^2$ over the interval $-5 < x < 1$, let’s examine a few key aspects:

1. Domain and Range:

• Domain: Given explicitly as $-5 < x < 1$.
• Range: To find the range, determine the maximum and minimum values of $f(x)$ on the interval $-5 < x < 1$.

2. Behavior of $f(x)$:

Since $f(x) = 2 - 6x^2$, this function is a downward-opening parabola (the coefficient of $x^2$ is negative), meaning it reaches a maximum at the vertex.

3. Vertex Calculation:

The vertex of a parabola $ax^2 + bx + c$ is located at $x = -\frac{b}{2a}$.

• Here, $a = -6$ and $b = 0$, so the vertex $x$-value is: $x = -\frac{0}{2 \cdot (-6)} = 0$
• Plugging $x = 0$ into $f(x)$: $f(0) = 2 - 6(0)^2 = 2$ So, $f(x)$ has a maximum value of 2 at $x = 0$.

4. Boundary Values:

Evaluate $f(x)$ at the boundaries of the domain:

• At $x = -5$: $f(-5) = 2 - 6(-5)^2 = 2 - 6 \cdot 25 = 2 - 150 = -148$
• At $x = 1$: $f(1) = 2 - 6(1)^2 = 2 - 6 = -4$

5. Range of $f(x)$:

Given that $f(x)$ reaches a maximum of 2 (at $x = 0$) and ranges down to -148 at $x = -5$, the range of $f(x)$ on $-5 < x < 1$ is: $-148 < f(x) \leq 2$

Summary

The function $f(x) = 2 - 6x^2$ over the interval $-5 < x < 1$ has:

• Domain: $-5 < x < 1$
• Range: $-148 < f(x) \leq 2$

Would you like further details on any step?

Related Questions:

1. How does the vertex form of a parabola affect its maximum and minimum values?
2. What would change in $f(x)$ if the interval were $-5 \leq x \leq 1$?
3. How does the negative coefficient in $-6x^2$ influence the shape of $f(x)$?
4. What is the process for finding critical points of $f(x)$ on a closed interval?
5. How would you determine concavity for different functions like $f(x) = 2 - 6x^2$?

Tip:

When analyzing the range of a function, always examine the function's critical points and endpoint behavior within the interval.