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

MATH110 April 2018

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Question 02 (b)
For what value of the constant ${\textstyle k}$ is the function ${\textstyle f}$ continuous everywhere? $f(x)={\begin{cases}1-kx&{\text{ if }}x<1\\kx^{2}-{\frac {2}{x}}&{\text{ if }}x\geq 1.\end{cases}}$

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Hint
When do we have ${\textstyle \lim _{x\to 1^{-}}f(x)=\lim _{x\to 1^{+}}f(x)}$ ?

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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 domain of the function $y=kx^{2}-{\frac {2}{x}}$ is $\{x\in \mathbb {\mathbb {R} } :x\neq 0\}$ , because any polynomials are defined on the real line and the rational function ${\frac {2}{x}}$ is defined on $\{x\in \mathbb {\mathbb {R} } :x\neq 0\}$ . However, the domain of the function $f$ is $\mathbb {R}$ because in order to compute $f(0)$ , we have to use the branch $1-kx$ as this one is defined for $x<1$ . (so $f(0)=1-k0=1$ ) On the domain, observe that for any value of ${\textstyle k}$ , the function will be continuous for all ${\textstyle x<1}$ and for all ${\textstyle x>1}$ , so the only point to check is ${\textstyle x=1}$ . The function will be continuous at ${\textstyle x=1}$ if and only if $\lim _{x\to 1^{-}}f(x)=\lim _{x\to 1^{+}}f(x),$ which by definition is equivalent to $\lim _{x\to 1}(1-kx)=\lim _{x\to 1}(kx^{2}-{\frac {2}{x}}).$ We have $\lim _{x\to 1}(1-kx)=1-k\cdot 1=1-k$ and $\lim _{x\to 1}(kx^{2}-{\frac {2}{x}})=k\cdot 1-{\frac {2}{1}}=k-2,$ so ${\textstyle f}$ will be continuous at ${\textstyle x=1}$ if and only if ${\textstyle 1-k=k-2}$ . After solving for ${\textstyle k}$ , we conclude that ${\textstyle f}$ is continuous everywhere if and only if ${\textstyle k=3/2}$ . Answer:* The function ${\textstyle f}$ is continuous everywhere if and only if ${\textstyle {\color {blue}k=3/2}}$ .