# $Let \; f(x) = \left\{\begin{array}{ll} -x^3 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \; , 0 \leq x < 1 \\ 2x - 3 \; , 1 \leq x \leq 3 \end{array} \right.$

Q: $\large Let \; f(x) = \left\{\begin{array}{ll} -x^3 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \; , 0 \leq x < 1 \\ 2x – 3 \; , 1 \leq x \leq 3 \end{array} \right.$

Find all possible real values of b such that f(x) has the smallest value at x = 1.

Sol. At x = 1, f(x) = -1

⇒ Smallest value of f(x) = -1

⇒ at all other points of the interval, f(x) > -1.

Now, for x ≥ 1, f(x) = 2x –3

⇒ f ‘(x) = 2 > 0 ⇒ f(x) is an increasing function

⇒ least value exists at x =1.

Again, for x < 1, f ‘(x) = -3x2 < 0

⇒ f(x) is decreasing function in the interval 0 ≤ x < 1.

Therefore, f(x) is smallest at x =1 provided

$\displaystyle f(1-0) = \lim_{h\rightarrow 0} -(1-h)^3 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq -1$

$\displaystyle -1 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq -1$

$\displaystyle \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq 0$

$\displaystyle \frac{(b^2 + 1)(b – 1}{(b + 1)(b +2 )} \geq 0$

$\displaystyle \frac{(b – 1}{(b + 1)(b +2 )} \geq 0$ ; Since (b^2 + 1) is always +ve

⇒ b ∈ (-2 ,-1) ∪ [ 1 , ∞)