Question

Complete the following statement.

Answer 1

\(\LARGE\color{black}{ f(x) = \begin{cases} & x^2+2,{~~~~}{\large x\le1} \\ & ax,{~~~~~~~~~~}{\large x>1} \end{cases} }\)

Answer 2

So, you want that \(x^2+2\), at \(x=1\),
will be equivalent to \(ax\), at \(x=1\).

Answer 3

How would I do that

Answer 4

\(\large\color{black}{ \displaystyle (1)^2+2=a(1) }\)

Answer 5

a times 1 is just 1.

and: (1)²+2=1+2=3

Answer 6

So the value of a that you want is equal to?

Answer 7

To the value in x^2 + 2?

Answer 8

Again, you want \(ax\) and \(x^2+2\) to be equal at x=1, so that there is no discontinuity at x=1.

Answer 9

a = 3?

Answer 10

\(\large x^2+2=ax\)
\(\large 1^2+2=a\cdot 1\)

\(\large {~~~~~~~}a~=~?\)

Answer 11

1^2+2=3
3 * 1 = 3
a = 3 right? :)

Answer 12

yes a=3

Answer 13

I didn't see that reply. I glitched. Again, yes you are right, a=3.

Answer 14

Thanks for teaching me :)

Answer 15

Not a problem