<p>Hi all I think following example of this topic contains an error.</p>
<p>_______<strong><em>quote from sparknote math II prepbook</em></strong>_____________
The Range of a Function with a Prescribed Domain
f(x) = 2x2 + 4 for –3 < x < 5. What is the range of f?
The best way to solve this type of problem is to manipulate the domain of x in exactly the same way that x is manipulated in the function. First x is squared, then multiplied by 2, then added to 4; we need to do the same thing to the bounds of the domain:</p>
<ol>
<li>–3 < x < 5</li>
<li>0 < x2 < 25 <----- isn't it 9?</li>
<li>0 < 2x2 < 50</li>
<li>4 < 2x2 + 4 < 54</li>
</ol>
<p>The range of f(x) is {4 < f(x) < 54}. </p>
<hr>
<p>Or do all negative numbers turn into 0 when squared in domain/range?
How hard is real math II test compared to sparknote test?</p>
<p>I got between 4 and 54 using calculus (don’t really know how to do it a different way). f(0)=4 and f(5)=54, therefore the range of f(x) for -3 < x < 5 is [4, 54].</p>
<p>I guess another way to think about it is because this is an “a*x^2 + b” function (where a and b are constants), f will have its largest value whenever |x| is the biggest (in this case, when x=5), and it will have the smallest value whenever x=0.</p>
<p>But to answer the main question, no, they did not make an error.</p>
<p>EDIT:
Okay, here is a better explanation, now that I’ve finally read through their steps.</p>
<ol>
<li> -3 < x < 5 <----- given</li>
<li> 0 < x^2 < 25 <----- x is between -3 and 5, so squaring 5 gets you 25, and squaring -3 does get you nine. HOWEVER, squaring -2, which is also in this range, gets you 4 (less than 9), -1 squared is 1, and 0 squared is 0, so the range so far is 0 to 25. Don’t forget about the x-values in between the end points. Zero squared is less than all positive AND negative numbers squared.</li>
<li>and 4. should be straightforward</li>
</ol>
<p>Hopefully that clears it up.</p>
<p>just plug in into your calculator, set the viewing window to -3 < x < 5, and then use the calculate the maximum and minumum</p>
<p>How did you do this with Calculus?</p>
<p>Well, right after I looked at the problem, calculus was just the first thing I thought of doing (basically just found absolute min, max) since that’s the class I’ve been taking the past two years. Then after I read through the entire thing I remembered how to do it the other way.</p>