CAN ANYONE SOLVE THIS STEP BY STEP

ALIAHMAD

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[imath]\sqrt{ \log_{ 0.4 }({ x- { x }^{ 2 } }) \phantom{\tiny{!}}}[/imath]
 
[imath]\sqrt{ \log_{ 0.4 }({ x- { x }^{ 2 } }) \phantom{\tiny{!}}}[/imath]
What is the question? What did you want to find?

Please post the complete problem - along with your attempt to solve it.
 
The honest answer to your question is no, nobody can solve this.
 
What is the question? What did you want to find?

Please post the complete problem - along with your attempt to solve it.
SIR PLEASE FIND THE DOMAIN OF THE ABOVE FUNCTION OR IN SIMPLE WORDS WHAT ARE THE VALUES OF X

The honest answer to your question is no, nobody can solve this.
I DIDN'T QUITE UNDERSTAND IT.
"SIR PLEASE FIND THE DOMAIN OF THE ABOVE FUNCTION OR IN SIMPLE WORDS WHAT ARE THE VALUES OF X"
PLEASE ADD THIS TO THE ABOVE QUESTION AND THEN PLEASE SOLVE IT
I REALLY NEED THE ANSWER
 
I REALLY NEED THE ANSWER
Please read the forum's guidelines. Here's a link to a summary.


Do you have any experience finding domains? For example, are you able to find the domain of this simpler version?

\(\displaystyle \sqrt{x - x^2}\)

Also, are you familiar with the change-of-base formula for logarithms?

[imath]\;[/imath]
 
Please read the forum's guidelines. Here's a link to a summary.


Do you have any experience finding domains? For example, are you able to find the domain of this simpler version?

\(\displaystyle \sqrt{x - x^2}\)

[imath]\;[/imath]
YES, I CAN SOVE IT THE DOMAIN OF THE FUNCTION [math]\sqrt{ x- { x }^{ 2 } \phantom{\tiny{!}}}[/math] IS x belongs to [0,1]

Please read the forum's guidelines. Here's a link to a summary.


Do you have any experience finding domains? For example, are you able to find the domain of this simpler version?

\(\displaystyle \sqrt{x - x^2}\)

Also, are you familiar with the change-of-base formula for logarithms?

[imath]\;[/imath]
YEAH LIKE IF THE BASE OF LOG IS 0<x<1 THEN WHILE REMOVING LOG THE SIGN OF INEQUALITY CHANGES
BUT IF THE BASE x>1 THEN SIGN OF INEQUALITY DOESN'T CHNAGE
 
I would set the log function part greater than or equal to 0 and find the domain . Because you can only take square root of positive values.
 
I would set the log function part greater than or equal to 0 and find the domain . Because you can only take square root of positive values.
I DON'T THINK THIS IS HOW LOG FUNCTIONS WORK WHEN GIVEN A CONSTANT BASE.
 
You have two constraints to consider.
1) You can't take the square root of a negative number (equivalent to what you take the square root of must be >=0)
2) You can only compute the log of a positive value.

log0.4(x-x2)> 0
x-x2>0
 
You have two constraints to consider.
1) You can't take the square root of a negative number (equivalent to what you take the square root of must be >=0)
2) You can only compute the log of a positive value.

log0.4(x-x2)> 0
x-x2>0
In this case I got the answer as x belongs to [0,1]
 
But the square root must also be >=0
Therefore x(x-1)>=-1
Which is x²-x+1>=0
Gives no answer as d=b²-4ac gives me negative value
 
You have [math]f(x) = g(h(i(x))), \text { where } i(x) = x - x^2,\\ h(x) = log_{0.4}(x), \text { and } g(x) = \sqrt{x}.[/math]
The way to do these problems is to start from the inside out.

What is the domain of i(x)? All real numbers.

What is the domain of h(x), a log function? What is the domain of any log function? [imath](0, \infty)[/imath].

So x values for which

[math]i(x) > 0 \implies x - x^2 > 0 \implies x > x^2.\\ x < 0 \implies 0 < x^2 \implies x < x^2 \implies x \not < 0.\\ x = 0 \implies x = x^2 \implies x \ne 0.\\ \therefore \ 0 < x.\\ x \ge 1 \implies x > 0 \implies x * x \ge x * 1 \implies x^2 \ge x \implies x < 1.\\ \text {In short, } 0 < x < 1.[/math]
Thus, the domain of f(x) cannot be LARGER THAN (0, 1). But it could be smaller. You have not completed the task.

Given x in (0, 1), what is the range of h(x)?

Is that larger than the domain of g(x).

If so, what next?
 
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