How Do You Solve Absolute Value and Quadratic Functions Together?

[takercena]

Homework Statement

a. Find the value of x such as fx < gx where fx = |2x -1| and gx = x(2-x)
b. evaluate \displaystyle\int^1_0 [gx - fx]\,dx

Homework Equations

none

The Attempt at a Solution

For question a I make it into 2 equation to 2x-1 = 2x-x^2 and 1 - 2x = 2x - x^2. I solve it and find the value of x = 1, 0.2679 and 3.73. The problem is, which interval should i choose if there is no graph to be sketched? And how do i get 0.2679 = 2 - sqrt of 3?

I have no idea for question b.

Thanks :)

 

[Dick]

You should always sketch a graph, whether they ask you to or not. It will make your life much easier. |2x-1|=2x-1 for x>=1/2 and |2x-1|=1-2x for x<=1/2. Work separately on those two intervals. There are only two points of intersection. And to get 0.2679...=2-sqrt(3) exactly, use the quadratic formula to solve the equation. Don't just put it into a calculator and get the numerical results.

 

[statdad]

Problem 1:
I haven't checked your numbers yet, but you shouldn't find three solutions, only 2. Once you have found the values of x where f(x) = g(x), those are the only places the two functions can be equal. Call the two locations of equality a, b, and for purposes of my notes assume that a &lt; b .&lt;br /&gt; &lt;br /&gt; Pick three numbers x_1 &amp;amp;lt; a, a &amp;amp;lt; x_2 &amp;amp;lt; b, and b &amp;amp;lt; x_3. Compare the function values at each x. If f &amp;amp;lt; g at your choice, it will be for all other values in that interval. (The same if g &amp;amp;gt; f).&lt;br /&gt; &lt;br /&gt; For problem 2: I assume you want&lt;br /&gt; &lt;br /&gt; &amp;lt;br /&amp;gt; \int_0^1 |g(x) - f(x)| \, dx&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; To do this you&amp;#039;ll need to split this integral into pieces, depending on where g(x) &amp;amp;gt; f(x) and g(x) &amp;amp;lt; f(x). But that&amp;#039;s exactly why you solved problem 1.