Question

4b) Determine the image sets \(g(J)\) for \(g(x) = x^{3}-3x^{2} + 1\) and the intervals
\(J:= [-1,3]\) and\( J:= [-3,1]\), respectively

Answer 1

\[
g'(x)=3 x^2-6 x=3 (x-2) x
\]
You can deduce that g has a local maximum at x=0 and a local minimum at x =2.
Both 0 and 2 are inside[-1,3]
\[
\{g(-1),g(0),g(2),g(3)\}=\{-3,
1,-3,1\}
\]
What can you conclude?

Answer 2

g([-1,3]=[-3,1]

Answer 3

\[
\{g(-1),g(0),g(-3)\}=\{-3,1,-5
3\}\\
\]
Hence
g[-3,1]=[-53,1]

Answer 4

THank you very much Mr Elias,

Answer 5

yw