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# Find the maximum value of 2x ^3 – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [–3, –1].

10. Find the maximum value of $2 x^3 - 24 x + 107$ in the interval [1, 3]. Find the
the maximum value of the same function in [–3, –1].

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Given function is
$f(x) = 2x^3-24x+107$
$f^{'}(x)=6x^2 - 24 \\ f^{'}(x)=0\\ 6(x^2-4) = 0\\ x^2-4=0\\ x^{2} = 4\\ x = \pm2$    we neglect the value x =- 2 because  $x \ \epsilon \ [1,3]$
Hence, x = 2  is the only  critical value  of function $f(x) = 2x^3-24x+107$
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 1 and x = 3
$f(2) = 2(2)^3-24(2)+107\\ = 2\times 8 - 48+107\\ =16-48+107 = 75$

$f(3) = 2(3)^3-24(3)+107\\ = 2\times 27 - 72+107\\ =54-72+107 = 89$

$f(1) = 2(1)^3-24(1)+107\\ = 2\times 1 - 24+107\\ =2-24+107 = 85$
Hence, maximum value of function  $f(x) = 2x^3-24x+107$  occurs at x = 3 and vale is 89  when $x \ \epsilon \ [1,3]$
Now, when $x \ \epsilon \ [-3,-1]$
we neglect the value x = 2
Hence, x = -2  is the only  critical value  of function $f(x) = 2x^3-24x+107$
Now, we need to check the value at x = -2 and at the end points of given range i.e. x = -1 and x = -3
$f(-1) = 2(-1)^3-24(-1)+107\\ = 2\times (-1) + 24+107\\ =-2+24+107 = 129$

$f(-2) = 2(-2)^3-24(-2)+107\\ = 2\times (-8) + 48+107\\ =-16+48+107 = 139$

$f(-3) = 2(-3)^3-24(-3)+107\\ = 2\times (-27) + 72+107\\ =-54+72+107 = 125$
Hence, the maximum value of function  $f(x) = 2x^3-24x+107$  occurs at x = -2 and vale is 139  when $x \ \epsilon \ [-3,-1]$

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