I need help please been getting all the answers wrong
2 answers:
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Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be 2 - 5i and:
Answer B.
Answer B.
F(x)=![\sqrt[3]{x+2}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%2B2%7D%20)
to solve for the inverse of a function you do 4 steps:
1. subsitute f(x) with y
2. switch y and x places
3. solve for y
4. subsitute y with f⁻¹(x)
so we have
f(x)=
subsitute f(x) with y
y=
switch x and y
x=![\sqrt[3]{y+2}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7By%2B2%7D%20)
solve for y
x=
cube both sides
=y+2
subtract 2 from both sides
=y
subsitute y with f⁻¹(x)
f⁻¹(x)=
the answer is f⁻¹(x)=
to solve for the inverse of a function you do 4 steps:
1. subsitute f(x) with y
2. switch y and x places
3. solve for y
4. subsitute y with f⁻¹(x)
so we have
f(x)=
subsitute f(x) with y
y=
switch x and y
x=
solve for y
x=
cube both sides
subtract 2 from both sides
subsitute y with f⁻¹(x)
f⁻¹(x)=
the answer is f⁻¹(x)=