Question

Find the extreme values of f(x)=Sqx+3

Answer 1

\[f(x) = \sqrt{x} ~~or~~ f(x) = \sqrt{x + 3}\]

Answer 2

the second one

Answer 3

It is square root and not a square, right?

Answer 4

\[\sqrt{x-3}\]

Answer 5

\[\sqrt{x+3}\], sory its plus

Answer 6

sqrt(x+3) is only defined for x >= -3.
It is a function that ALWAYS increases as x increases.
Therefore, can you guess where it will have a minimum?

Answer 7

i dont get the second line

Answer 8

Put x = -3, -2, -1, 0, 1, 2, 3, etc. and you can see the value of sqrt(x+3) will keep on increasing. That is what I meant in the second line.

Answer 9

so would this be the case for every equation with a square root ?

Answer 10

It depends on the expression inside the square root.
Here, inside the square root, you have x+3 which keeps on increasing as x increases and therefore sqrt(x+3) will also keep on increasing.
But if you have some other expression, then it may increase and then decrease, etc. and so sqrt will also behave accordingly.

Answer 11

like what if it was x-3 ?

Answer 12

x -3 is also a function that keeps on increasing as x increases. If you plot y = x - 3, you will see it is a straight line with a positive slope meaning it keeps on increasing as x increases. So sqrt(x-3) will also keep on increasing.

Answer 13

sqrt(1000 - x), for example, will keep on DECREASING as x increases.

Answer 14

The function, x^2 - 8x + 30 keeps decreasing for x < 4 and keeps increasing for x > 4.
So sqrt(x^2 - 8x + 30) will also keep decreasing for x < 4 and keeps increasing for x > 4.

Answer 15

so what would the solution be to the promblem ?

Answer 16

a0 When x = -3 then (x - 3) = 0
b) when x = +infinity.
If x = -infinity, the function is undefined.