Calculate the Descending Area of f(x) = (1/2)x²: Quadratic Function Analysis

Q: What's the difference between where a function is negative and where it's decreasing?

A function being negative means f(x) decreasing means it's going downward as x increases, which happens when f'(x)

Q: Why do I need to find the derivative to determine where it's decreasing?

The derivative f'(x) tells you the slope at each point. When f'(x)

Q: How do I know if f'(x) = x is correct?

Use the power rule: $ \frac{d}{dx}(\frac{1}{2}x^2) = \frac{1}{2} \cdot 2x^{2-1} = \frac{1}{2} \cdot 2x = x $

Q: What does 'descending area' mean exactly?

Descending area means the same as decreasing interval - the domain where the function's values get smaller as x increases from left to right.

Q: Can a parabola like this have more than one decreasing interval?

No! For $ f(x) = \frac{1}{2}x^2 $, there's exactly one decreasing interval (x 0), separated by the vertex at x = 0.

Quadratic Functions with Derivative Analysis

Find the descending area of the function

$f(x)=\frac{1}{2}x^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the domain of decrease of the function

00:03 Notice the coefficient of X squared is positive, so the function is concave up

00:07 Let's look at the trinomial coefficients

00:15 Use the formula to find the vertex of the parabola

00:21 Substitute appropriate values according to the given data and solve to find the vertex

00:24 This is the X value at the vertex of the parabola

00:29 Determine when the parabola decreases and increases based on its type

00:33 Draw the X-axis and find the domain of decrease

00:42 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the descending area of the function

$f(x)=\frac{1}{2}x^2$

Step-by-step solution

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: $x < 0$ .

Final Answer

$x < 0$

Key Points to Remember

Essential concepts to master this topic

Rule: Function decreases where its derivative is negative
Technique: Find f'(x) = x, then solve x < 0
Check: Test point x = -1: f'(-1) = -1 < 0, so decreasing ✓

Common Mistakes

Avoid these frequent errors

Confusing where function is negative with where it's decreasing
Don't look at where f(x) < 0, look at where f'(x) < 0! The function f(x) = (1/2)x² is always positive, but it decreases when x < 0. Always use the derivative to find increasing/decreasing intervals.

Practice Quiz

Test your knowledge with interactive questions

Which chart represents the function $$ y=x^2-9 $$ ?

FAQ

Everything you need to know about this question

What's the difference between where a function is negative and where it's decreasing?

A function being negative means f(x) < 0 (below the x-axis). A function decreasing means it's going downward as x increases, which happens when f'(x) < 0.

Why do I need to find the derivative to determine where it's decreasing?

The derivative f'(x) tells you the slope at each point. When f'(x) < 0, the slope is negative, meaning the function is going downward (decreasing).

How do I know if f'(x) = x is correct?

Use the power rule: $\frac{d}{dx}(\frac{1}{2}x^2) = \frac{1}{2} \cdot 2x^{2-1} = \frac{1}{2} \cdot 2x = x$

What does 'descending area' mean exactly?

Descending area means the same as decreasing interval - the domain where the function's values get smaller as x increases from left to right.

Can a parabola like this have more than one decreasing interval?

No! For $f(x) = \frac{1}{2}x^2$ , there's exactly one decreasing interval (x < 0) and one increasing interval (x > 0), separated by the vertex at x = 0.

How can I visualize this without graphing?

Think of the parabola opening upward with vertex at (0,0). As you move left from the vertex (x < 0), you're going downhill, so it's decreasing!

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