Question

Consider \[f(x) = {x^3} + a{x^2} + bx + c,\forall x \ge 0\], where parameters a, b, c are chosen respectively by throwing a dice three times. Then the probability that f(x) is an increasing function is .

Answer 1

Hint: If the derivative \[f'(x)\] of a continuous function \[f(x)\] satisfies \[f'(x) > 0\] on an open interval \[(a,b)\] , then \[f(x)\] is increasing on \[(a,b)\] . However, a function may increase on an interval without having a derivative defined at all points. For example, the function \[{x^{\dfrac{1}{3}}}\] is increasing everywhere, including the origin \[x = 0\] , despite the fact that the derivative is not defined at that point.

Complete step by step answer:
We are given that the terms a, b, c are chosen by throwing a dice simultaneously now we know that the numbers may appear in a dice are 1, 2, 3, 4, 5, 6
Which means that all the numbers are positive thus the domain for a,b,c is \[\{ 1,2,3,4,5,6\} \]
Now If we try to find the derivative of \[f(x) = {x^3} + a{x^2} + bx + c\]
We will get \[f'(x) = 3{x^2} + 2ax + b\]
Which clearly means that \[f'(x) > 0,\forall x \ge 0\] because a and b both are positive quantity and for x to be the least that is when x is equal to 0 then also we will have \[f'(0) = b\]
And if \[x \ge 1\] we will always have a positive quantity
Thus we can say that the function \[f(x)\] is always increasing
Hence the probability is 1.

Note:
Probability is simply how likely something is to happen. Whenever we're unsure about the outcome of an event, we can talk about the probabilities of certain outcomes—how likely they are. It can have values from 0 to 1, 0 if and only if there's no chance of an event to occur and 1 if it is sure that the event will occur.