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].

Answers (1)

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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