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Question 515247: find the maximum value of the function
f(x)= 3+4x^2-x^4
hint: let t=x^2
Found 2 solutions by drcole, richard1234:
Answer by drcole(72)   (Show Source):
You can put this solution on YOUR website! The first thing to notice here is that f(x) is a continuous function, and that, if we took its limit as x approaches  or  , the function approaches because of the coefficient of the leading term ( ) is -1 and goes to  in either direction. So this function will not have an absolute minimum (it can be as negative as we want), but because it is continuous everywhere and approaches in either direction, it WILL have an absolute maximum.
We also notice that this function has a continuous derivative everywhere, since it is a polynomial. If we know that a function has a continuous derivative everywhere, and we know that it has an absolute maximum, that maximum must be at a stationary point: a point where the first derivative equals zero. So we need to find the derivative, set it equal to zero, and solve for x.
 (applying the power rule for differentiation)
 (simplifying)
 (setting the first derivative equal to zero)
 (dividing both sides by -2)
 (factoring out an x)
 (factoring using the difference of two squares)
Setting each factor equal to 0, we get the roots  ,  , and  . These are the only possible x-values for the location of the maximum of f(x). We now plug each of these roots into f(x) and find the largest value among the three for the maximum.
So the maximum value of f(x) is 7, and it occurs at two values of x: and .
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! We can maximize
 = \LARGE 3 + 4t - t^2 = -t^2 + 4t + 3)
The vertex occurs at t = -(4/-2) = 2. If t = 2, then we can replace t = 2 (or x = sqrt(2)) to obtain 7 as the maximum value.
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