Science:Math Exam Resources/Courses/MATH110/April 2018/Question 02 (e)

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MATH110 April 2018

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Question 02 (e)
Find the absolute minimum of ${\textstyle g(x)={\frac {1}{x}}+\ln(x)}$ .

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
If $g$ has a extremum at $x=a$ , then $a$ is a critical point. So, find critical points of $g$ .

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Solution
By the Hint, the critical points are candidates for the points at which an extremum occurs. Recall that a function $g$ has a critical number $x=a$ in the domain if it satisfies one the the following conditions. $g'(a)=0$ , $g'(a)$ does not exist a is an end point. So, we find critical points of $g$ . Observe that $g={\frac {1}{x}}+\ln x$ has a domain $\{x\in \mathbb {R} :x>0\}$ (since the logarithmic function can only take positive arguments), and on the domain the given function $g$ is differentiable everywhere. In other words, there's no point at which $g'(x)$ doesn't exist. Also, we can see that our domain doesn't have end point because it is an open interval $(0,+\infty )$ . Now, we find the roots of the first derivative of ${\textstyle g}$ . The first derivative of ${\textstyle g}$ is given by $g'(x)=\left({\frac {1}{x}}+\ln(x)\right)'=\left({\frac {1}{x}}\right)'+\left(\ln(x)\right)'=-{\frac {1}{x^{2}}}+{\frac {1}{x}}.$ Setting this to zero and rearranging gives ${\frac {1}{x^{2}}}={\frac {1}{x}}.$ By multiplying through by ${\textstyle x^{2}}$ , we obtain ${\textstyle x=1}$ . Therefore, $x=1$ is the only critical point of $g$ . It remains to check that this is indeed a global minimum. Indeed, it suffices to show that for a single ${\textstyle x\neq 1}$ (in ${\textstyle (0,+\infty )}$ ), one has ${\textstyle g(x)>g(1)=1}$ . If ${\textstyle x=e}$ , then $g(e)={\frac {1}{e}}+\ln(e)={\frac {1}{e}}+1>1,$ as required. Answer: The function ${\textstyle g(x)}$ has a global minimum at ${\textstyle {\color {blue}x=1}}$ .