If the function f given by f(x) = x3 -- 3(a -- 2)x2 + 3ax + 7, for some a[ ]R is increasin…

Question

If the function f given by f(x) = x3 -- 3(a -- 2)x2  + 3ax + 7, for some a$$ \in $$R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation,  $${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$$ is :

Accepted Answer

f '(x) = 3x2 - 6(a - 2)x + 3a

f '(x) $$ \ge $$ 0 $$\forall $$ x $$ \in $$ (0, 1]

f '(x) $$ \le $$ 0 $$\forall $$ x $$ \in $$ [1, 5)

$$ \Rightarrow $$  f '(x) = 0  at  x = 1 $$ \Rightarrow $$  a = 5

f(x) - 14 = (x - 1)2 (x - 7)

$${{f(x) - 14} \over {{{\left( {x - 1} \right)}^2}}} = x - 7$$

Solutions · 1