Science:Math Exam Resources/Courses/MATH110/April 2011/Question 08 (a)

MATH110 April 2011

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Question 08 (a)
Let $\displaystyle f(x)=\ln \left(4-x^{2}\right)$ Find the domain of ƒ.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Where is the function $\ln(x)$ defined? What does that tell you about the argument ($4-x^{2}$) of this function?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The function $\ln(x)$ is defined for all $x>0$. Therefore, in order for $\ln(4-x^{2})$ to be defined, we need $4-x^{2}>0$. If we factor the left side of the inequality, we get $\displaystyle (2-x)(2+x)>0$ For what values of x is this inequality true? If $x=2$ or -2, then the left side of the inequality is zero, which is not allowed. So 2 and -2 cannot be part of our domain. Similarly, if $x>2$ or $x<-2$, then the left side of the inequality will be negative, contradicting the inequality. So these values are also excluded from our domain. If we choose $-2<x<2$ we find that the left hand side of the inequality is positive, satisfying the inequality. Thus our domain is $-2<x<2$, which written in interval notation is $\displaystyle (-2,2)$.

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