[Calculus] Continuous Functions (1 Viewer)

youngsky · May 12, 2013

Having trouble with this question. I have the answers but they're not the same as mine :/

"If the continuous function h(x) is defined by h(x) =

1. 4x-6 ; x <= -1
2. ax^2 + bx ; -1 < x< 2
3. x^3 ; x >= 2

Find a and b."

D94 · May 13, 2013

Is the answer, a=14 and b=24?

SpiralFlex · May 13, 2013

$I will assume$\; h \;$was defined as a piece-wise function$

$For a function$ \;f,\; $to be continuous we need,$

$\lim_{x\rightarrow a}f(x)=f(a)$

$$Well in simpler terms, I want the left hand limit to equal to the right hand limit if they exist to both equal to the value of$ \;f\; $at$\; a.$

$\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a)$

$$In our case we observe we need to investigate the limits for case$\; x \rightarrow -1, x \rightarrow 2.\; $It is clear we need to construct two equations solving our two unknowns$\; a,b.$

[HR][/HR]

$$Case I:$\; x \rightarrow -1$

$\lim_{x\rightarrow -1^+}h(x)=a(-1)^2+b(-1)=a-b$

$\lim_{x\rightarrow -1^-}h(x)=4(-1)-6=-10$

$h(-1)=4(-1)-6=-10$

$$We require,$\; \lim_{x\rightarrow -1^+}h(x)=\lim_{x\rightarrow -1^-}h(x)=h(-1)$

$a-b=-10.....(1)$

[HR][/HR]

$$Case II:$\; x \rightarrow 2$

$\lim_{x\rightarrow -2^+}h(x)=(2)^3=8$

$\lim_{x\rightarrow -2^-}h(x)=a(2)^2+b(2)=4a+2b$

$h(2)=4(2)-6=2$

$$We require,$\; \lim_{x\rightarrow 2^+}h(x)=\lim_{x\rightarrow 2^-}h(x)=h(2)$

$4a+2b=2.....(2)$

$2a+b=1$

[HR][/HR]

$$Solving$\; (1)\; $and$\; (2)\; $simultaneously we attain$\; a=-3, b=7$

youngsky · May 13, 2013

SpiralFlex said:
$I will assume$\; h \;$was defined as a piece-wise function$

$For a function$ \;f,\; $to be continuous we need,$

$\lim_{x\rightarrow a}f(x)=f(a)$

$$Well in simpler terms, I want the left hand limit to equal to the right hand limit if they exist to both equal to the value of$ \;f\; $at$\; a.$

$\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a)$

$$In our case we observe we need to investigate the limits for case$\; x \rightarrow -1, x \rightarrow 2.\; $It is clear we need to construct two equations solving our two unknowns$\; a,b.$

[HR][/HR]

$$Case I:$\; x \rightarrow -1$

$\lim_{x\rightarrow -1^+}h(x)=a(-1)^2+b(-1)=a-b$

$\lim_{x\rightarrow -1^-}h(x)=4(-1)-6=-10$

$h(-1)=4(-1)-6=-10$

$$We require,$\; \lim_{x\rightarrow -1^+}h(x)=\lim_{x\rightarrow -1^-}h(x)=h(-1)$

$a-b=-10.....(1)$

[HR][/HR]

$$Case II:$\; x \rightarrow 2$

$\lim_{x\rightarrow -2^+}h(x)=(2)^3=8$

$\lim_{x\rightarrow -2^-}h(x)=a(2)^2+b(2)=4a+2b$

$h(2)=4(2)-6=2$

$$We require,$\; \lim_{x\rightarrow 2^+}h(x)=\lim_{x\rightarrow 2^-}h(x)=h(2)$

$4a+2b=2.....(2)$

$2a+b=1$

[HR][/HR]

$$Solving$\; (1)\; $and$\; (2)\; $simultaneously we attain$\; a=-3, b=7$

Worked it out myself in the end

Book's answers were a = -2, b = 8

Also SpiralFlex in your last line of working 4a + 2b = 8, but otherwise it looks great

Thanks bro

SpiralFlex · May 14, 2013

My mistake well spotted

[Calculus] Continuous Functions (1 Viewer)

youngsky

poof

D94

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SpiralFlex

Well-Known Member

youngsky

poof

SpiralFlex

Well-Known Member

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