Parabola of the Form y=x²+c: Finding Increasing or Decreasing Domains

To determine the intervals where the function $f(x) = 2x^2$ is increasing, we will analyze the derivative of the function:

Step 1: Differentiate the function.
The derivative of $f(x) = 2x^2$ is $f'(x) = 4x$.

Step 2: Determine where $f'(x) > 0$.
To find the increasing intervals, set $4x > 0$. Solving this inequality, we obtain $x > 0$.

Therefore, the function $f(x) = 2x^2$ is increasing for $x > 0$.

Consequently, the correct answer is the interval where the function is increasing, which is $0 < x$.

To solve the problem of finding the descending area of the function $f(x) = \frac{1}{2}x^2$, we follow these steps:

• Step 1: Calculate the derivative of the given function. The function is $f(x) = \frac{1}{2}x^2$. Differentiating this, we get $f'(x) = \frac{d}{dx}(\frac{1}{2}x^2) = x$.
• Step 2: Determine where the derivative is negative. Since $f'(x) = x$, the derivative is negative when $x < 0$.
• Step 3: Conclude the solution. We find that the function $f(x)$ is decreasing for $x < 0$.

Thus, the descending area (domain where the function is decreasing) for the function $f(x) = \frac{1}{2}x^2$ is $x < 0$.

The correct choice that matches this solution is: $x < 0$.

To solve this problem, we'll identify the intervals where the function $f(x) = -2x^2 + 10$ is increasing and decreasing. Here’s how we can tackle it:

• Step 1: Locate the Vertex
  Since the function is $f(x) = -2x^2 + 10$, it follows the standard form of a quadratic: $ax^2 + bx + c$ where $a = -2$, $b = 0$, and $c = 10$. The vertex of this parabola lies at $x = -\frac{b}{2a}$. Plugging the values in, we find: \begin{align*} x &= -\frac{0}{2(-2)} \\ x &= 0 \end{align*} So, the vertex is located at $x = 0$.
• Step 2: Determine Increasing and Decreasing Intervals
  Since $a = -2$, which is negative, the parabola opens downward. This means the function will increase as it approaches the vertex and decrease after passing it. Therefore: \begin{enumerate}
• The function is increasing when $x < 0$.
• The function is decreasing when $x > 0$.
• Verification with Choices
  Comparing with the given choices, the correct answer is that the function is decreasing when $x > 0$ and increasing when $x < 0$.

Therefore, the intervals are:

$0 < x$ decreasing

$x < 0$ increasing

To determine the ascending area of the function $f(x) = 6x^2 - 12$, we will follow these steps:

• Step 1: Calculate the derivative of the given function.
• Step 2: Set the inequality $f'(x) > 0$ to find the interval where the function is increasing.
• Step 3: Solve the inequality for $x$.

Let's begin with Step 1:
The derivative of $f(x) = 6x^2 - 12$ with respect to $x$ is:

$f'(x) = \frac{d}{dx}(6x^2 - 12) = 12x$.

Step 2: We need to find where $12x > 0$. This requires:

$x > 0$.

Step 3: Therefore, the function $f(x) = 6x^2 - 12$ is increasing when $x > 0$.

Thus, the increasing interval of the function is when $x > 0$.

The solution to the problem is $0 < x$.

To solve this problem, the focus is on determining the increasing intervals of the function $f(x) = -4x^2 - 24$.

Here's how we'll proceed:

• Step 1: Differentiate the function.
• Step 2: Set the derivative greater than 0 to determine the intervals where the function ascends.
• Step 3: Analyze the results to determine the intervals of increase.

Step 1: Find the derivative of $f(x) = -4x^2 - 24$.
The derivative is a straightforward calculation:
$f'(x) = \frac{d}{dx}(-4x^2 - 24) = -8x$.

Step 2: Solve $f'(x) > 0$ to find the increasing interval.
$-8x > 0$ leads to $x < 0$.

Step 3: Conclude by analyzing this result.
This tells us that the function is increasing when $x < 0$, meaning the ascending area of $f(x)$ lies in this interval.

Therefore, the solution to this problem is $x < 0$.

Let's solve the problem.

Step 1: Calculate the derivative of the function:

The function is $f(x) = -3x^2 + 12$.

The derivative $f'(x)$ is calculated using the power rule:

$f'(x) = \frac{d}{dx}(-3x^2 + 12) = -6x$.

Step 2: Find where the derivative is positive:

To find the interval where the function is increasing, solve the inequality $f'(x) > 0$.

This yields:

$-6x > 0$

Divide both sides by $-6$ (remember to reverse the inequality sign when dividing by a negative):

$x < 0$

Therefore, the function is increasing for $x < 0$.

Thus, the ascending area of the function is $x < 0$.

To determine the increasing and decreasing domains of the quadratic function $f(x) = 5x^2 - 25$, we begin by analyzing its structure:

This function is a quadratic function of the form $ax^2 + c$. Here, $a = 5$, which is positive. As such, the parabola opens upwards.

The vertex of such a quadratic function, when $b = 0$, is simply at $x = 0$. Thus, the symmetry point of the parabola is based on this vertex.

Since the parabola opens upwards:

• The function is decreasing on the interval $(-\infty, 0)$.
• The function is increasing on the interval $(0, \infty)$.

Therefore, the function $f(x) = 5x^2 - 25$ is:

$x < 0$ decreasing

$0 < x$ increasing

Thus, the correct answer choice for the intervals is the one provided in Choice 4.

To determine where the function $f(x) = 2x - x^2 + 1$ is decreasing, we follow these steps:

• Step 1: Find the derivative
  The derivative of the function $f(x) = 2x - x^2 + 1$ is found using standard calculus rules. Thus, $f'(x) = \frac{d}{dx}(2x - x^2 + 1) = 2 - 2x$.
• Step 2: Determine where the derivative is negative
  Set the derivative $2 - 2x$ less than zero to find where the function is decreasing:

$2 - 2x < 0$

Solve for $x$:
Subtract 2 from both sides to get:

$-2x < -2$

Now, divide both sides by -2, remembering to reverse the inequality sign:

$x > 1$

Conclusion: The function $f(x) = 2x - x^2 + 1$ is decreasing for $x > 1$. Thus, the descending area is represented by the interval $1 < x$.

The correct choice that matches this interval is:

$1 < x$