# Critical Number from Analying a Graph.

• 1calculus1
In summary, the homework statement says that you need to find all critical numbers of f(x)= x\sqrt{2x+1}. The equations show that the function is differentiable there, so -1/2 is one of the critical points. If x > -1/2, then you can solve for dy/dx = 0 and find x = -1/3.

## Homework Statement

Find all critical numbers of f(x)= x$$\sqrt{2x+1}$$

## Homework Equations

Derivative, Product Rule

## The Attempt at a Solution

x$$\sqrt{2x+1}$$ = 0
x must be -1/2 & 0
(-1/2, 0) Critical Point.
(0, 0) Critical Point.

First derivative:
f'g + g'f
(1)($$\sqrt{2x+1}$$)+(1/2$$\sqrt{2x+1}$$)(x)(2)
=($$\sqrt{2x+1}$$) + (x / $$\sqrt{2x+1}$$)
LCD:
(($$\sqrt{2x+1}$$) ($$\sqrt{2x+1}$$) + 1)) / ($$\sqrt{2x+1}$$)

= 2x+1 / $$\sqrt{2x+1}$$
So y= 0 when x= -1/2 and 0
(-1/2, 0) Critical Point.

The second derivative I wouldn't post it her since it gets pretty messy. However, I found out that y=o for the second derivative must have x = -1/2

1calculus1 said:

## Homework Statement

Find all critical numbers of f(x)= x$$\sqrt{2x+1}$$

## Homework Equations

Derivative, Product Rule

## The Attempt at a Solution

x$$\sqrt{2x+1}$$ = 0
x must be -1/2 & 0
(-1/2, 0) Critical Point.
(0, 0) Critical Point.

First derivative:
f'g + g'f
(1)($$\sqrt{2x+1}$$)+(1/2$$\sqrt{2x+1}$$)(x)(2)
=($$\sqrt{2x+1}$$) + (x / $$\sqrt{2x+1}$$)
LCD:
(($$\sqrt{2x+1}$$) ($$\sqrt{2x+1}$$) + 1)) / ($$\sqrt{2x+1}$$)
$y=x \sqrt{2x+1}$

$$\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{2\sqrt{2x+1}}(2)$$

$$\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{\sqrt{2x+1}}$$

$$\frac{dy}{dx}=\frac{(\sqrt{2x+1})^2 + x}{\sqrt{2x+1}}=0$$

rock.freak667 said:
$y=x \sqrt{2x+1}$

$$\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{2\sqrt{2x+1}}(2)$$

$$\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{\sqrt{2x+1}}$$

$$\frac{dy}{dx}=\frac{(\sqrt{2x+1})^2 + x}{\sqrt{2x+1}}=0$$

OH! Thanks for that!
But, the thing is.. the answer for the critical number is -1/3 and from looking at the first derivative, one of the critical point is -1/2.

So what am I doing wrong?

While the derivative is undefined at x = -1/2, is the original function differentiable there, and what does that tell you about whether -1/2 is a critical number? Otherwise, if x > -1/2, then you can solve for dy/dx = 0, and you should find x = -1/3. Try it again using the derivative rock.freak posted.

Tedjn said:
While the derivative is undefined at x = -1/2, is the original function differentiable there, and what does that tell you about whether -1/2 is a critical number? Otherwise, if x > -1/2, then you can solve for dy/dx = 0, and you should find x = -1/3. Try it again using the derivative rock.freak posted.

OH! Thanks for that. Problem solved.

## What is a critical number in graph analysis?

A critical number in graph analysis is a point on a graph where the derivative is equal to zero or does not exist. It indicates a potential change in the direction of the graph, such as a maximum or minimum point.

## How do you find critical numbers on a graph?

To find critical numbers on a graph, you can take the derivative of the function and set it equal to zero. Then, solve for the variable to find the x-values of the critical numbers. Alternatively, you can visually inspect the graph for any points where the slope is zero or undefined.

## Can a critical number be a point of inflection?

No, a critical number cannot be a point of inflection. A point of inflection is a point on a graph where the second derivative changes sign, indicating a change in concavity. A critical number only indicates a potential change in the direction of the graph, but it does not necessarily indicate a change in concavity.

## What information can critical numbers provide about a graph?

Critical numbers can provide information about the maximum and minimum points of a graph. They can also help identify any potential points of inflection. Additionally, critical numbers can be used to determine the intervals where a function is increasing or decreasing.

## Do all functions have critical numbers?

No, not all functions have critical numbers. A function must be differentiable and continuous for critical numbers to exist. For example, a function with a vertical asymptote will not have any critical numbers.