Question 515247
The first thing to notice here is that f(x) is a continuous function, and that, if we took its limit as x approaches {{{+infinity}}} or {{{-infinity}}}, the function approaches {{{-infinity}}} because of the coefficient of the leading term ({{{x^4}}}) is -1 and {{{x^4}}} goes to {{{infinity}}} 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 {{{-infinity}}} 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.

{{{ f(x) = 3 + 4x^2 - x^4 }}}
{{{ (df(x))/(dx) = 0 + 4(2x) - 4x^3 }}} (applying the power rule for differentiation)
{{{ (df(x))/(dx) = 8x - 4x^3 }}} (simplifying)
{{{ 0 = 8x - 4x^3 }}} (setting the first derivative equal to zero)
{{{ 0 = x^3 - 2x }}} (dividing both sides by -2)
{{{ 0 = x(x^2 - 2) }}} (factoring out an x)
{{{ 0 = x(x + sqrt(2))(x - sqrt(2)) }}} (factoring using the difference of two squares)

Setting each factor equal to 0, we get the roots {{{x = 0}}}, {{{x = sqrt(2) }}}, and {{{x = -sqrt(2)}}}.  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.

{{{ f(0) = 3 + 4(0^2) - 0^4 = 3}}}
{{{ f(sqrt(2)) = 3 + 4((sqrt(2))^2) - (sqrt(2))^4 = 3 + 4(2) - 4 = 3 + 8 - 4 = 7 }}}
{{{ f(-sqrt(2)) = 3 + 4((-sqrt(2))^2) - (-sqrt(2))^4 = 3 + 4(2) - 4 = 3 + 8 - 4 = 7 }}}

So the maximum value of f(x) is 7, and it occurs at two values of x: {{{x = sqrt(2) }}} and {{{x = -sqrt(2)}}}.