y=2x2-3x2-12x+5=0
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Find the maximum and minimum values that the function can take over all the values in the domain for the input. The range is the maximum minus the minimum.
Set the first derivative of the function equal to zero, and solve for the variable.
The general procedure is to find the function's derivative, and then solve for (derivative of the function) = 0. Each of these solutions may be a local maximum or minimum - or none. Further analysis is required. A local maximum or minimum may also occur at points where the derivative is undefined, as well as at the function's endpoints (assuming it is only defined for a certain range, for example, from 0 to 10).
the maximum or minimum value of a continuous function on a set.
A quadratic function can only have either a maximum or a minimum point, not both. The shape of the graph, which is a parabola, determines this: if the parabola opens upwards (the coefficient of the (x^2) term is positive), it has a minimum point; if it opens downwards (the coefficient is negative), it has a maximum point. Therefore, a quadratic function cannot exhibit both extreme values simultaneously.