Answer: \frac{(x+5)}{3} Step-by-step solution, Assume f\left(x\right)=y So y=3x-5 and x=\frac{\left(y+5\right)}{3} So x=\frac{\left(f\left(x\right)+5\right)}{3} So inverse function is \frac{\left(x+5\right)}{3}
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What is the global maximum and minimum values of the function f(x) = 2x^3 - 3x^2 + 1 on the interval [-1,2]?
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