# f(x)=9, if x<=-4, and f(x)=ax+b, if -4<x<5, and f(x)=-9, if x>=5,Find

$$\displaystyle{f{{\left({x}\right)}}}={9},{\quad\text{if}\quad}{x}\le-{4},{\quad\text{and}\quad}{f{{\left({x}\right)}}}={a}{x}+{b},{\quad\text{if}\quad}-{4}{<}{x}{<}{5},{\quad\text{and}\quad}{f{{\left({x}\right)}}}=-{9},{\quad\text{if}\quad}{x}\ge{5},$$
Find the constants, if the function is continuous on the entire line.

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sweererlirumeX
The function is continuous, if
$$\displaystyle{\underset{{{x}\to{a}^{+}}}{{\lim}}}{f{{\left({x}\right)}}}={\underset{{{x}\to{a}^{{-}}}}{{\lim}}}{f{{\left({x}\right)}}}={f{{\left({a}\right)}}}$$
All of the given functions are continous, since $$\displaystyle{\underset{{{1}}}{{{f}}}}{\left({x}\right)},{\underset{{{3}}}{{{f}}}}{\left({x}\right)}$$ are constant function and $$\displaystyle{\underset{{{2}}}{{{f}}}}{\left({x}\right)}$$ is linear function. We need to check only at point where the function might break (-4 and 5):
Atx=-4
$$\displaystyle{\underset{{{x}\to-{4}^{{-}}}}{{\lim}}}{f{{\left({x}\right)}}}={\underset{{{x}\to-{4}^{{-}}}}{{\lim}}}{9}$$
=9
$$\displaystyle{\underset{{{x}\to-{4}^{+}}}{{\lim}}}{f{{\left({x}\right)}}}={\underset{{{x}\to-{4}^{{-}}}}{{\lim}}}{a}{x}+{b}$$
=a(-4)+b
=-4a+b
f(-4)=9
$$\displaystyle{\underset{{{x}\to-{4}^{{-}}}}{{\lim}}}{f{{\left({x}\right)}}}={\underset{{{x}\to-{4}^{+}}}{{\lim}}}{f{{\left({x}\right)}}}={f{{\left(-{4}\right)}}}$$
9=-4a+b=9
-4a+b=9 (I)
At x=5
$$\displaystyle{\underset{{{x}\to{5}^{+}}}{{\lim}}}{f{{\left({x}\right)}}}={\underset{{{x}\to{5}^{{-}}}}{{\lim}}}{a}{x}+{b}$$
=5a+b
$$\displaystyle{\underset{{{x}\to{5}^{+}}}{{\lim}}}{f{{\left({x}\right)}}}={\underset{{{x}\to{5}^{+}}}{{\lim}}}-{9}$$
=-9
f(5)=-9
If the condition of continuity is satisfied,
$$\displaystyle{\underset{{{x}\to{5}^{{-}}}}{{\lim}}}{f{{\left({x}\right)}}}={\underset{{{x}\to{5}^{+}}}{{\lim}}}{f{{\left({x}\right)}}}={f{{\left({5}\right)}}}$$
5a+b=-9 (II)
Subtract (II) from (I) to get:
-9a+0=18
$$\displaystyle\to{a}=-{2}$$
Substitute a=-2 in (I)
-4(-2)+b=9
8+b=9
b=1
a=-2
b=1