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December 18, 2014

December 18, 2014

Posted by **Mishaka** on Saturday, October 22, 2011 at 9:07pm.

f(x) = {3x^2 - 2x +1, if x < 0

a cos(x) + b, if 0 </= x </= pi/3

4sin^2(x), if x > pi/3

A. 0

B. 1

C. 2

D. 3

E. 4

I know that since the function is continuous, it should be equal to 1 at 0 and 3 at pi/3 (To follow the other two pieces of the function). From here, I am having a great deal of difficulty figuring out what coordinates would make the function work in this way. Any help is appreciated.

- Calculus (Continuity) -
**Reiny**, Saturday, October 22, 2011 at 10:03pmTo be continuous, the functions should have the same value at the transition values of x

First transition value: x = 0

at x = 0, 3x^2 - 2x + 1 = 1

and acosx + b = acos(0) + b = a+b

since they wanted the value of a+b and we know its value is 1

we are done: a+b = 1

check:

at x= π/3

acosπ/3 + b = a/2 + b

and 4sin^2 (π/3) = 4(√3/2)^2 = 4(3/4) = 3

then a/2 + b = 3

a+2b = 6

solving with a+b=1, subtract them

b = 5

then a+5 = 1

a = -4

so when x=0 , first function is 1

2nd function is -4cos0 + 5 = -4+5 = 1 , good

when x=π/3

2nd function is -4cos(π/3) + 5 = -4(1/2) + 5 = 3

3rd function is 4sin^2 (π/3) = 4(√3/2)^2 = 3 , good!

So the values of a=-4 and b=5

or

a+b=1

make the functions continuous.

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