Let a function fx-x33-2x2-3x on the set A-x- x2-8- 6x- then the minimum value of f is

Given : nbsp;f(x)=x^33 2x^2+3x ⇒ f'(x)=x^2 4x+3=(x 1)(x 3) ⇒ f'(x)=0, gives x=1,3 now, set A={x|x^2 6x+8≤0} ⇒set A=[2,4] So, critical points for f(x) is x=2,3 ...

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Standard XII

Mathematics

Monotonicity in an Interval

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Question

Let a function $f (x) = \frac{x^{3}}{3} - 2 x^{2} + 3 x$ on the set $A = {x | x^{2} + 8 \leq 6 x},$ then the minimum value of $f$ is

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Solution

Given : $f (x) = \frac{x^{3}}{3} - 2 x^{2} + 3 x$
$\Rightarrow f^{'} (x) = x^{2} - 4 x + 3 = (x - 1) (x - 3)$
$\Rightarrow f^{'} (x) = 0,$ gives $x = 1, 3$
now, set $A = {x | x^{2} - 6 x + 8 \leq 0}$
$\Rightarrow set A = [2, 4]$
So, critical points for $f (x)$ is $x = 2, 3, 4$ and
$f (2) = \frac{2}{3} f (3) = 0 f (4) = \frac{4}{3}$
So, $f_{min} = 0$

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Monotonicity in an Interval

Standard XII Mathematics

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Let a function f(x)=x33−2x2+3x on the set A={x|x2+8≤6x}, then the minimum value of f is

Given : f(x)=x33−2x2+3x ⇒f′(x)=x2−4x+3=(x−1)(x−3) ⇒f′(x)=0, gives x=1,3 now, set A={x|x2−6x+8≤0} ⇒setA=[2,4] So, critical points for f(x) is x=2,3,4 and f(2)=23f(3)=0f(4)=43 So, fmin=0

Let a function $f (x) = \frac{x^{3}}{3} - 2 x^{2} + 3 x$ on the set $A = {x | x^{2} + 8 \leq 6 x},$ then the minimum value of $f$ is