MHB 004- USMA(WP) entrance exam question independent variable

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The independent variable of the function W(θ) = 2θ² is restricted to the interval [2, 6]. Consequently, the dependent variable ranges from 8 to 72, as W(θ) is an increasing function for positive values of θ. The calculations confirm that for θ values in the specified range, W(θ) yields results between 8 (when θ=2) and 72 (when θ=6). There is some clarification regarding the interpretation of θ, emphasizing that it is simply a number and not necessarily an angle measurement. Overall, the discussion focuses on the relationship between the independent and dependent variables in the context of the given function.
karush
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ok not sure what forum this was supposed to go in,,so...

If the independent variable of $W(\theta)=2\theta^2$ is restricted to values in the interval [2,6]
What is the interval of all possible values of the dependents variable?
 
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[8, 72]
 
For positive values of \theta, \theta^2, and therefore 2\theta^2, is an increasing function. Values of 2\theta^2, for x between 2 and 6, lie between 2(2^2)= 8 and 2(6^2)= 72.
 
HallsofIvy said:
For positive values of \theta, \theta^2, and therefore 2\theta^2, is an increasing function. Values of 2\theta^2, for x between 2 and 6, lie between 2(2^2)= 8 and 2(6^2)= 72.

Ok I think I got ? Because $\theta$ ussually means radians or degrees!
 
Nothing is said, in this problem, about \theta being the measure of an angle or any measurement at all. It's just a number.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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