We need to find the least possible value of a and the greatest possible value of b.
We know that the curve:
[tex]y=x³-2x²+4x-9[/tex]
has no stationary points in the interval a < x
This means that y'(x) ≠ 0 for any x in the interval between a and b.
Thus, first, let's find the points where y'(x) = 0:
[tex]\begin{gathered} y^{\prime}(x)=0 \\ \\ 3x²-2\cdot2x+4=0 \\ \\ 3x²-4x+4=0 \\ \\ x=\frac{-(-4)\pm\sqrt{(-4)²-4(3)(4)}}{2(3)} \\ \\ x=\frac{4\pm\sqrt{-32}}{6} \end{gathered}[/tex]
Since the values of x for which y'(x) = 0 are imaginary, there are no real values of x at which the given curve is stationary.
Therefore, a and b can be any real number.
Answer:
The minimum value of a is represented by -∞, and the maximum value of b is represented by ∞.
