Find Values of a & b

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Calculus
Section 2.5

How is this done algebraically?

Screenshot_20220522-083844_Samsung Notes.jpg
 
to make f continuous, you need to find intersection points of all three given functions:

start with first function
(x^2-4)/(x-2)=x+2

second function is: ax^2-bx+3

equal it to first function:
ax^2-bx+3=x+2..........solve for a
ax^2=bx-3+x+2
ax^2=bx+x-1
a=(bx+x-1)/x^2.........eq.1

do same using third function:
2x-a+b=x+2......solve for a
2x-x+b-2=a
a=x+b-2...........eq.2

from eq.1 and eq.2 we haveL

(bx+x-1)/x^2=x+b-2 ......solve for b
b (x^2 - x) = -x^3 + 2 x^2 + x - 1
b = (-x^3 + 2 x^2 + x - 1)/(x^2 - x)

go to eq/2, substitute b
a=x+ (-x^3 + 2 x^2 + x - 1)/(x^2 - x) -2

a=(x^2 - 3 x + 1)/(x - x^2)

now we have a and be expressed in terms of x

so, second function will be
ax^2-bx+3=((x^2 - 3 x + 1)/(x - x^2))x^2-( (-x^3 + 2 x^2 + x - 1)/(x^2 - x) )x+3=x+2
=>a=0, b=-1
third function will be
2x-a+b=2x-0-1=2x-1

so,
ax^2-bx+3=x+2
2x-a+b=2x-0-1=2x-1

3160-247256644516b66ef6b08f4fe1df022f.jpg


MSP3663147c9ddi2h92341f000061ei48831bg2g8hd
is continuous
 

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to make f continuous, you need to find intersection points of all three given functions:

start with first function
(x^2-4)/(x-2)=x+2

second function is: ax^2-bx+3

equal it to first function:
ax^2-bx+3=x+2..........solve for a
ax^2=bx-3+x+2
ax^2=bx+x-1
a=(bx+x-1)/x^2.........eq.1

do same using third function:
2x-a+b=x+2......solve for a
2x-x+b-2=a
a=x+b-2...........eq.2

from eq.1 and eq.2 we haveL

(bx+x-1)/x^2=x+b-2 ......solve for b
b (x^2 - x) = -x^3 + 2 x^2 + x - 1
b = (-x^3 + 2 x^2 + x - 1)/(x^2 - x)

go to eq/2, substitute b
a=x+ (-x^3 + 2 x^2 + x - 1)/(x^2 - x) -2

a=(x^2 - 3 x + 1)/(x - x^2)

now we have a and be expressed in terms of x

so, second function will be
ax^2-bx+3=((x^2 - 3 x + 1)/(x - x^2))x^2-( (-x^3 + 2 x^2 + x - 1)/(x^2 - x) )x+3=x+2
=>a=0, b=-1
third function will be
2x-a+b=2x-0-1=2x-1

so,
ax^2-bx+3=x+2
2x-a+b=2x-0-1=2x-1

3160-247256644516b66ef6b08f4fe1df022f.jpg


MSP3663147c9ddi2h92341f000061ei48831bg2g8hd
is continuous

What can I say? Simply amazing!!
 

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