Verified answer

If you're allowed some calculus, the first deriative f '(x) = 2 +2x

and the second derivative f "(x) is 2, implying that the critical values

( -1) for the first derivative gives a relative min in your interval of

definition: f(-1) = zero, a relative min (which is an absolute min too).

That the function increases left and right of -1 implies a max/rel-max

at the endpts; you check f(-2) = +1 and f(2) = +9. Thus you have your

range: [0, +9]

[You could describe f(-2) as a rel-max.]

You can find the range by entering the points of the range into the function, when:

x=-2 f(x)=1, (-2+1)^2=-1^2=1

x=-1 f(x)=0, (-1+1)^2=0^2=0

x=0 f(x)=1, (0+1)^2=-1^2=1

x=1 f(x)=4, (1+1)^2=-2^2=4

x=2 f(x)=9, (2+1)^2=3^2=9

When in doubt just enter the parameters to find the bounds of the equation.

hmmm...probable...however the only that is going terrific for the wizard or the only he's terrific suitable for stands out as the only chosen. that's no longer likely although that the point of compatibility between the two wands would be realtively extreme! One is limited to be greater properly matched. additionally, each and each wand has a center, and a particular lentgh. The length of a wand relies upon on the dimensions of the wizards arm, and the middle finally relies upon on what variety of magic the wizard prefers. If a wizard is greater apt for Transfiuration , as an occasion, then that man or woman would be greater valuable to a particular middle. and likewise for sure if the wand is won, then that wand provides you with its allegiance...making you an proprietor of two wands...very resembling Harry and Dumbledore! So i think of its achieveable, yet exceedingly no longer likely! And it very infrequently takes place (i think of)

Where did u get those choices from? Textbook incorrect?

I dont see a way to get zero thats the point... o.O