Question

if f(x)=(x^-x -x^0 + 2^x), then f(3) is equal to

Answer 1

Plug in "3" wheerever you see an \(\sf \color{Red}{x}\)

Answer 2

ya i got that, but what am i supposed to do to the -x

Answer 3

first of all.
\[x^0=1\]as any number to the power of zero. (Ask me to prove if you want to know)

Also,\[\color{red}{ a^{-b}= \frac{1}{a^b} } \]

Answer 4

Prove that s**t, bro!

Answer 5

\[\huge\color{red}{ x^0=x^{a-a}=x^a \div x^a=\frac{x^a}{x^a} } \]

Answer 6

@abb0t, good?

Answer 7

f(3)=(3/3)+(1)+(2^3)=10

Answer 8

x^-x is same as x/x which is 3/3=1

Answer 9

\[ f(x)=(x^{-x} -x^0 + 2^x)\]\[x^0=1~~~~~~~So,\]\[ f(x)=(x^{-x} -1 + 2^x)\]\[ f(x)=(\frac{1 }{x^x} -1 + 2^x)\]

go so far?

Answer 10

x^0=1

Answer 11

solomon is right x^-x=1/(3^3)=1/9

Answer 12

i meant 1/27

Answer 13

I am solving what you wrote.

If you want to find the zeros, set it equal zero.

\[\frac{1 }{x^x}-1+2^x=0\]\[\frac{1 }{x^x}+2^x=1\]\[\frac{1 }{x^x}+\frac{2^x~x^x }{x^x}=1\]\[\frac{1 +2^xx^x}{x^x}=1\]\[Log(\frac{1 +2^xx^x}{x^x})=Log(1)\]

so far so good?

Answer 14

ur answer is (1/27)+1+8=244/27

Answer 15

the answer choice are
1) 8 1/27
2) -21
3) 7 1/27
4) -22

Answer 16

Yes, if you plug in 3, but don't you want to know the zeros, if you don;t plug in 3?

Answer 17

\[ f(x)=(x^ {-x} -x^0 + 2^x)~~~~~~~~~ f(x)=(3^ {-3} -3^0 + 2^3)~~~~~~~~~ f(x)=(\frac{1 }{27} -1 +9)\]

Answer 18

\[ f(x)=\frac{1 }{27} +8~~~~~~ \]

Answer 19

its (1/27)-(1)+(8)

Answer 20

My bad, yeah

Answer 21

2^3=8 not 9 solomon

Answer 22

so its 7+ 1/27

Answer 23

its (1/27)-9

Answer 24

thankss!!

Answer 25

amrita the answer is 3) 190/27 which is 7 1/27 the way u get that is (1/(3^3))-1+8=(1/27)-1+8=(1/27)+7=190/27 so answer is 3)