Question

Find a so that f(x)= {ax^2-9 if x 2 is continuous at x=2

Answer 1

this is much easier than it seems.
replace \(x\) by 2 in both expressons, set them equal, solve for \(a\)

Answer 2

\[f(x) = \left\{\begin{array}{rcc}
ax^2-9& \text{if} & x \leq 2 \\
4x+3& \text{if} & x > 2
\end{array}
\right.
\]

Answer 3

you get
\[a\times 2^2-9=4\times 2+3\]
\[4a-9=11\] etc

Answer 4

but don't i need to set both to a limit ?

Answer 5

yes of course, but both of these are continuous functions, so the limit is what you get when you replace \(x\) by 2

Answer 6

the idea is that you are trying to match them up so they are equal where the function changes definition

Answer 7

in other words, they should agree when \(x=2\)
when \(x=2\) the first one is \(4a-9\) and you want to make sure it agrees with the second one, which would be \(11\)

Answer 8

i got a = 9/4

Answer 9

you will notice that the first thing is said was "it is easier than it seems"
it really is just elementary algebra right?

Answer 10

i don't think that is right
did you solve
\[4a-9=11\] ?

Answer 11

i don't understand why you are making 4a-9 equal to 11

Answer 12

ok lets go slow

Answer 13

when you replace \(x\) by 2 in the top formula \(ax^2-9\) what do you get?

Answer 14

4a-9

Answer 15

ok good. now you know that \(ax^2-9\) is a polynomial, so the limit is identical to the value of the function at any number because all polynomials are continuous everwhere

Answer 16

so the value at \(x=2\) is \(4a-9\)
now what do you get when you replace \(x\) by 2 in the second formula?

Answer 17

so the lim as x->2- ax^2-9 = 2?

Answer 18

in the seconand formula it = 11

Answer 19

no not 2, but rather \(f(2)\)

Answer 20

the limit is the same as the value of the function, not the same as \(x\)

Answer 21

\[\lim_{x\to 2}ax^2-9=a\times 2^2-9=4a-9\]

Answer 22

so \[\lim x \rightarrow0 of ax ^{2}-9=f(2)\]

Answer 23

and \(\lim_{x\to 2}4x+3=4\times 2+3=11\)

Answer 24

right, thats what i have.

Answer 25

ok so the limit from the left is \(4a-9\) and the limit from the right is \(11\) and you want the function to be continuous, so you have to make sure those limits are the same. ergo set
\[4a-9=11\] and solve for \(a\)

Answer 26

a=5

Answer 27

yup

Answer 28

so f(x)=11=f(2)=a ?

Answer 29

now lets go back to the beginning as see how easy this was and what needed to be done
replace \(x\) by 2 in both expressions, set them equal, solve for \(a\)

Answer 30

what is \(a\) ?

Answer 31

5

Answer 32

so what is \(ax^2-9\) ?

Answer 33

11

Answer 34

when u sub in 5

Answer 35

no, but rather when you replace \(a\) by 5 in the expression \(ax^2-9\) ?

Answer 36

i think there is some confusion here. \(a\) is just some constant, the coefficient of \(x^2\) for the first formula
now you know \(a=5\) because if \(a=5\) then the top formula becomes \(5x^2-9\) and the bottom formula is \(4x+3\) and if you replace \(x\) by 2 in either formula, you get \(11\) in both cases, meaning the limit is the same from the left and from the right
you answer is
\[f(x) = \left\{\begin{array}{rcc}
5x^2-9& \text{if} & x \leq 2 \\
4x+3& \text{if} & x > 2
\end{array}
\right.\]

Answer 37

you can see from the answer i wrote, that if \(x=2\) the first one gives \(5\times 4-9=11\) and the second one gives \(4\times 2+3=11\) so both limits are the same, and the function is continuous now

Answer 38

so i have to re-write my function with the 5 instead of the a?

Answer 39

yes, the whole question was to find out what \(a\) should be

Answer 40

if \(a\) was not 5, say \(a=3\) then the function would not be continuous at 2, because the limit would not be 11, but it would still be 11 for the second formula

Answer 41

okay thank-you,

Answer 42

yw