Maximum and Minimum of a cubic function

Question

Maximum value of function

$y = x^3-5x^2+2$

a) 5
b) $\infty$
c) 2
d) -5

We know to find maximum value of a function we take first derivative of the function and make it zero and get some point. And checking second derivative of it. But in this case similar way we get $x=0$ has maximum and maximum value will be 2 but the maximum value of the function not 2 . It will be infinite if we check the graph of the function. But how can mathematically we can say the maximum value of the function will be infinite?

Answer 1

There is a slight mistake in your argument: taking the first derivative, making it vanish, and then checking the second derivative is a recipe to find the local maxima/minima of the function, not the global maxima/minima.

Finding the global maxima is an altogether different task. You have to collect the local maxima and compare them in order to find the global maxima(which will work only if the function is bounded). Mostly, one must find global maxima using some rough sketch of the graph.

Your function is a cubic polynomial, namely:$y = x^3 - 5x^2 + 2$

You can make a rough sketch as follows:

1. Calculate the first derivative and make it vanish. This implies:

    $3x^2 - 10x = 0$
   

   Thus, we get x=3.33 and 0 as the points of extrema

2. Calculate the second derivative and check the sign of it at the points of extrema. Here, the second derivative is:

    6x-10
   

   We get negative sign at x = 0 and positive sign at x = 3.33

   Thus, from negative infinity, the function rises till x=0, sees a small dip till x=3.33 and then rises upwards till infinity.

   Another explanation for this is that x^3 dominates over x^2 and x: it can be seen clearly that for values greater than 1: $x<x^2<x^3$. Hence, the graph of the function will be dominated by x^3 and be taken to infinity.

   Attached here is a graph of the function for reference:

   https://www.desmos.com/calculator/yg8h18wvuz