Question

Determine the values of a and b for which the following function is continuous at every real number x

f(x) { ax+b if x ≤ 1
-2bx^2 - 8a if 1 < x < 2
4b^2/a -5x if x ≥ 2 }

Answer 1

Can someone please help me ? :(

Answer 2

\[f(x)=\begin{cases} (1) ax+b & , x\leq 1\\(2) -2bx^2-8a &,1<x<2\\(3) 4b^2/a-5x &, 2\leq x \end{cases}\]
let \(x=1\) and set equation (1) = equation (2)
let \(x=2\) and set equation (2) = equation (3)

solve the two equations.

Answer 3

@sirm3d

Answer 4

isn't it a(1)+b = -2a(1)^2 -8a ?

Answer 5

my mistake.
\[a(1)+b=-2b(1)^2-8a\\a+b=-2b-8a\\3b=-9a\\b=-3a\]

Answer 6

equating (2) and (3) when \(x=2\) and \(b=-3a\)

\[-2(-3a)(2)^2-8a=\frac{4(-3a)^2}{a}-5(2)\\24a-8a=36a-10\\16a=36a-10\\10=20a\\\frac{1}{2}=a\]

Answer 7

You could use limits also.

Answer 8

can you show me how to do it using limits ? @abb0t

Answer 9

@alyannahere Are you here?

Answer 10

yes i am @ash2326 pls help me. and most preferably use limits :)

Answer 11

Yes, it's easy to solve this using limits,
\[f(x)=\begin{cases} (1) ax+b & , x\leq 1\\(2) -2bx^2-8a &,1<x<2\\(3) 4b^2/a-5x &, 2\leq x \end{cases}\]

For this Let's find the limit at x=1, for f(x) to be continuous the left hand and right hand limit should be equal

Answer 12

Do you get this part?

Answer 13

yes. so it's going to be

ax+b = a + b
x->1^+

ax+b = -a + b
x->1^_

right ?

Answer 14

For left hand limit use ax +b x<=1
For right hand limit use -2bx^2-8a 1<x<2
Both should be equal

Answer 15

Just put x=1 and equate the two limits

Answer 16

a=0 ?

Answer 17

No, you can't find a from this, you need one more equation
For that find the limits at x=2 ( the equate the left and right hand limit )

Answer 18

i'm so sorry but i don't get you @ash2326

Answer 19

:(

Answer 20

Limit at x=1
Left hand limit
(ax+b),
put x=1
a+b
Right hand limit
-2bx^2-8a
put x=1
-2b-8a
For the function to be continous
\[a+b=-2b-8a\]
Do you get this part?

Answer 21

yup! and that's what i used to solve a but apparently you said it still has to be equal.. which i don't get already

Answer 22

This is first equation we need one more equation to find a and b

Now let's find limit at x=2
Left hand limit
-2bx^2-8a
put x=2
-8b-8a

Right hand limit
4b^2/a-5x
put x=2
\[4b^2/a-10\]
Now equate the two
\[-8b-8a=4b^2/a-10\]
Now use this and the previous equation to solve for a and b, Can you solve this now ?

Answer 23

okay i'll solve it and get back to you @ash2326 :D

Answer 24

Cool :D

Answer 25

a= 1/2
b= 3/2 ?

Answer 26

then after i get these... what do i do now ? ...to get the continuity

Answer 27

@ash2326

Answer 28

i am getting A=1/2 AND B=-3/2

Answer 29

sorry use a and b (small)

Answer 30

@alyannahere

Answer 31

oh yeah since it's b=-3a

Answer 32

would you know how to do continuity @nitz ?

Answer 33

oops my bad. b=-3/2

Answer 34

see we have determined the values of a and b assuming that the function is continuous at x=1 and x=2....that is why we have substituted x=1 in first two equations and x=2 in last two......because of continuity
\[\lim_{x \rightarrow 1-}f(x)=\lim_{x \rightarrow 1+}f(x)\]
ie left and right hand limits are equal

Answer 35

then ? so it's all equal?at a=1/2 b=-3/2 ... just like that ?

Answer 36

ya ....you were to find the values of a and b ....and that we have got

Answer 37

oh thank you ! @nitz

Answer 38

you are welcome....any other help ..

Answer 39

yes we are having a text on continuity tom and i don't seem to understand the concept.

Answer 40

which concept...

Answer 41

find the largest union of intervals at which the following functions is continuous. Show complete algebraic solution

{ |x+2| x<=0
3 0<x<=2
squareroot of (3x-15) x>2

Answer 42

continuity @nitz

Answer 43

it is discontinuous at x=0 and x=2

Answer 44

why ?

Answer 45

discontinuous at x=0 and x=2 because left hand and right hand limits are not equal...

Answer 46

\[\lim_{x \rightarrow 0-}f(x)=\lim_{x \rightarrow 0-}\left| x+2 \right|=\lim_{h \rightarrow 0}\left| (0-h)+2 \right|=\lim_{h \rightarrow 0}\left| -h+2 \right|=\lim_{h \rightarrow 0}\ \left| 0+2 \right|=2\]

Answer 47

\[\lim_{x \rightarrow 0+}f(x)=3\]

Answer 48

so.left hand limit is 2 and right hand limit is 3 which are not equal and hence the function is discontinuous at x=0

Answer 49

on a similar basis,....left hand limit at x=2 is 3 and right hand limit is 3(2)-15=-9 ,,,so again both are not equal and function is discontinuous at x=2 also

Answer 50

right?/

Answer 51

@alyannahere

Answer 52

so ....the function is continuous at all points except x=0 and x=2

Answer 53

R-{0,2}

Answer 54

are we not going to consider squareroot of (3x-15) x>2 ?

Answer 55

sorry i forgot the squareroot..

Answer 56

lol :))

Answer 57

but still when we check x=2 for continuity ...left hand limit ie for x<=2 the value of function is 3 and for right hand limit sqrt(3x-15)=sqrt(3(2)-15)=sqrt(6-15)=sqrt(-9)=3i....still as 3 is not equal to 3i,....so left and right hand limits are not equal and hence function is discontinuous at x=2

Answer 58

any doubt

Answer 59

no i think i understand it already. thanks @nitz !

Answer 60

you are welcome....for any help u can refer these tutorials
http://tutorial.math.lamar.edu/Classes/CalcI/Continuity.aspx