Let's analyze the function provided in the image:

Function:

$f(x) = 2x^2 + 2$

Interval:

The function is defined over the interval $-2 \leq x \leq 5$.

Steps to find the range:

1. Since this is a quadratic function $2x^2 + 2$, the graph will be a parabola opening upwards (because the coefficient of $x^2$ is positive).
2. To find the range, we need to check the values of $f(x)$ at the boundaries of the interval and the vertex of the parabola, since the vertex will give us the minimum value (as the parabola opens upwards).

Finding the vertex:

The vertex of a parabola in the form $f(x) = ax^2 + bx + c$ is given by the formula: $x_{vertex} = -\frac{b}{2a}$ For $f(x) = 2x^2 + 2$, $a = 2$ and $b = 0$, so: $x_{vertex} = -\frac{0}{2(2)} = 0$ Thus, the vertex occurs at $x = 0$.

Evaluating the function at key points:

• At $x = -2$: $f(-2) = 2(-2)^2 + 2 = 2(4) + 2 = 8 + 2 = 10$
• At $x = 5$: $f(5) = 2(5)^2 + 2 = 2(25) + 2 = 50 + 2 = 52$
• At the vertex, $x = 0$: $f(0) = 2(0)^2 + 2 = 2$

Conclusion:

• The minimum value of the function occurs at $x = 0$ and is $f(0) = 2$.
• The maximum value of the function occurs at $x = 5$ and is $f(5) = 52$.

Range:

Thus, the range of $f(x)$ over the interval $-2 \leq x \leq 5$ is: $2 \leq f(x) \leq 52$

Answer:

The correct choice based on the options provided is: $10 \leq f(x) < 52$

Would you like more details or have any questions?

Here are some related questions to explore:

1. How do you determine the direction of a parabola?
2. What is the effect of increasing the coefficient of $x^2$ on the range of the function?
3. How can we find the vertex of a parabola if the $b$-term is non-zero?
4. What are the general properties of quadratic functions?
5. How does changing the interval affect the range of a quadratic function?

Tip: Always evaluate the function at the endpoints and vertex when finding the range of quadratic functions over a closed interval.