Minima and Least Value

Calculus Level 5

Let f ( x ) = { ∣ x 2 − 3 x ∣ + a ; 0 ≤ x < 3 2 − 2 x + 3 ; x ≥ 3 2 f\left( x \right) =\begin{cases} \left| { x }^{ 2 }-3x \right| +a\quad ;\quad 0\le x<\frac { 3 }{ 2 } \\ -2x+3\quad ;\quad x\ge \frac { 3 }{ 2 } \end{cases} If f ( x ) f\left( x \right) has a local m a x i m a maxima at x = 3 2 x=\frac { 3 }{ 2 } , then the l e a s t least value of ∣ 4 a ∣ \left| 4a \right| is

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The answer is 9.00.

Utkarsh Bansal by Utkarsh Bansal , 23, India

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1 solution

Maxima occurs when x = 3 2 x=\frac32 . Substituting in f ( x ) f(x) , we get the local maxima as 0. This means ∣ x 2 − 3 x ∣ + a ≤ 0 ; x ∈ [ 0 , 3 2 ) |x^2-3x|+a\leq0 ; x \in [0,\frac32) .

∣ x 2 − 3 x ∣ |x^2-3x| is an increasing function in the range [ 0 , 3 2 ) [0,\frac32) , therefore Minimum value of a would be maximum value of x x . Since it is an increasing function, maxima is at lim ⁡ x → 3 2 ∣ x 2 − 3 x ∣ = 9 4 \lim_{x \rightarrow \frac 32} |x^2-3x| = \frac94 . Substituting a = − 9 4 ⇒ ∣ 4 a ∣ = 9 a= \frac{-9}{4} \Rightarrow |4a| = 9 .

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