The Notes: Find values for `a` and `b` that will make `f` continuous everywhere, if f(x) = { `-3x^2` if `x lt 1;` `ax + b` if `1lt=xlt=4;` `x + sqrt(x)` if `4...

Monday, December 28, 2009

Find values for $\displaystyle{a}$ and $\displaystyle{b}$ that will make $\displaystyle{f{}}$ continuous everywhere, if f(x) = { $\displaystyle-{3}{{x}}^{{2}}$ if $\displaystyle{x}\lt{1};$ $\displaystyle{a}{x}+{b}$ if $\displaystyle{1}\leq{x}\leq{4};$ $\displaystyle{x}+\sqrt{{{x}}}$ if $\displaystyle{4}\ldots$

Hello!

The formulas used to define $\displaystyle{f{{\left({x}\right)}}}$ are elementary functions and they are continuous on a given intervals. The only problem problematic points are the joint points $\displaystyle{1}$ and $\displaystyle{4}.$

Even at these points $\displaystyle{f{}}$ has limits from the left and from the right. And for $\displaystyle{f{}}$ to be continuous at $\displaystyle{1}$ and $\displaystyle{4}$ it is necessary and sufficient for these limits to coincide. It is evident that

$\displaystyle\lim_{{{x}\to{1}-}}{f{{\left({x}\right)}}}=-{3},$ $\displaystyle\lim_{{{x}\to{1}+}}{f{{\left({x}\right)}}}={a}+{b},$

$\displaystyle\lim_{{{x}\to{4}-}}{f{{\left({x}\right)}}}={4}{a}+{b},$ $\displaystyle\lim_{{{x}\to{4}+}}{f{{\left({x}\right)}}}={6}.$

This gives the linear system for $\displaystyle{a}$ and $\displaystyle{b},$ $\displaystyle{a}+{b}=-{3}$ and $\displaystyle{4}{a}+{b}={6}.$ Solve it by substitution: $\displaystyle{b}=-{3}-{a}$ and $\displaystyle{4}{a}-{3}-{a}={6},$ thus $\displaystyle{a}={3}$ and $\displaystyle{b}=-{6}.$ This is the (unique) answer.