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The difference of absoulte maximum (largest) and absoulte minimum (smallest) values of the function f(x)=12x4/3−6x1/3,x∈[−1,1]f(x) = 12x^{4/3} - 6x^{1/3}, x \in [-1, 1] is

Solution

✅ Correct Option: 4

f′(x)=16x1/3−2x−2/3=2x−2/3(8x−1)f'(x) = 16x^{1/3} - 2x^{-2/3} = 2x^{-2/3}(8x-1), so critical points are x=1/8x=1/8 and x=0x=0 (where f′f' fails to exist).

Values: f(−1)=18f(-1) = 18, f(0)=0f(0)=0, f(1/8)=−9/4f(1/8) = -9/4, f(1)=6f(1) = 6.

Difference =18+9/4=81/4= 18 + 9/4 = 81/4.

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