Question

2 Calculus questions

Answer 1

on the first question. you have to think about restrictions with radicals.

Answer 2

if you have anything less than 1 under the radical what happens

Answer 3

idk...

Answer 4

so what do you do with say.... sqrt-10

Answer 5

make it positive

Answer 6

?

Answer 7

remember anything about imaginary numbers from algebra?

Answer 8

An Imaginary Number, when squared, gives a negative result

An imaginary number is a number that can be written as a real number multiplied by the imaginary unit

Answer 9

well, where i am going. is there is a restriction where x cannot be less than 1 because you don't want negatives in the radical.

Answer 10

we cannot have a negative value under the square root sign or we will end up with a complex number.

Answer 11

correct. so which way is the graph continuous

Answer 12

right?

Answer 13

yes.

Answer 14

ok because numbers to the right of the graph are positive right?

Answer 15

is that for both 5 and 6?

Answer 16

@lonnie455rich

Answer 17

the next one is going to be a little differentt. think about the restrictions on x in the denominator. what will make it undefined?

Answer 18

The rational expression is undefined when the denominator equals to zero

Answer 19

alright. so what happens if you plug 1 in for x?

Answer 20

it's equals 1

Answer 21

uhh. g(1). what does it equal.

Answer 22

0

Answer 23

so what if you plug 0 into f(x)

Answer 24

Infinity?

Answer 25

uhh. no what happens if you have 0 in the denominator at a point on the graph? what happens to the graph

Answer 26

it crosses the y axis?

Answer 27

umm. it will have a point of discontinuity.

Answer 28

k.

Answer 29

do you understand why? because at that point x can not be defined on the graph

Answer 30

so it's not continuous at that point but why not?

Answer 31

because at x=1 of (f(g(x)) it is undefined

Answer 32

because h(x) is the composition of f and g of x
and its talking about the graph of h(x)