Question

Explain why (–4x)^0 = 1, but –4x^0 = –4

Answer 1

-4x^0 = -4 times x^0 = -4 times 1 = -4

anything to the 0 (except 0 and infinity) is 1. Calculus deals with the 0 and infinity cases.

so (-4x)^0 is like a^0 = 1, where a = -4x

Answer 2

I am confused.

Answer 3

How do you know that a=-4x?

Answer 4

(-4x)^3 = -64x^3
(-4x)^2 = 16x^2
(-4x)^1 = -4x^1

if we increase the exponent by 1 we multiply by -4x.

likewise, if we decrease the exponent by 1 we divide by -4x.

what is -4x divided by -4x?

Answer 5

1?

Answer 6

yep.

does that one makes sense?

Answer 7

Oh I think I get it now, if it was 4x^0 then it would be 1, right?

Answer 8

not quite... when written like that, x is the only thing being raised to the exponent.

4x^2 = 4*x*x, only the x is exponentiated.

do you remember the order of operations? (PEMDAS)?

exponents come first them multiplication. and that's what is going on there.

4 is a multiplier and x is the base.

with (-4x)^0 the base is -4x, the whole thing.

Answer 9

What do you mean the whole thing?

Answer 10

i mean -4x is the base. so (-4x)^2 = (-4x)*(-4x) = 16x^2

when you have -4x^2 = -4*x^2=-4*x*x

Answer 11

i mean the -4x is the base, not just the x. that's what i meant by the whole thing. hopefully that's a little clearer.

Answer 12

hmm, I think I get it know... Thanks, you earned yourself a medal and fan :)

Answer 13

thanks for your patience!!!

Answer 14

\(\large {\bf ({\color{brown}{ -4x}})^0\to {\color{brown}{ 1}}
\\ \quad \\
-4{\color{brown}{ x}}^0\to -4\cdot {\color{brown}{ x}}^0\to -4\cdot {\color{brown}{ 1}}\to -4 }\)