The minimum value of f(4) - f(1) is 6.
The maximum value of f(4) - f(1) is 12.
In the question, we are given that, 2 ≤ f'(x) ≤ 4 for all values of x.
Taking the given inequality as (i).
We are asked to find the minimum and maximum possible values of f(4) - f(1).
We multiply (i) by dx throughout, to get:
4dx ≤ f'(x)dx ≤ 5dx.
To find this, we integrate (i) in the definite interval [4, 1] with respect to dx, to get:
[tex]\int_{1}^{4}2dx \leq \int_{1}^{4}f'(x)dx \leq \int_{1}^{4}4dx\\\Rightarrow [2x]_{1}^{4} \leq [f(x)]_{1}^{4} \leq [4x]_{1}^{4}\\\Rightarrow 2*4 - 2*1 \leq f(4)-f(1) \leq 4*4 - 4*1\\\Rightarrow 6 \leq f(4) -f(1) \leq 12[/tex]
Thus, the minimum value of f(4) - f(1) is 6.
The maximum value of f(4) - f(1) is 12.
Learn more about definite integrals at
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