Given p(x)=x^4+ax^3+bx^2+cx+d,such that x=0 is the only real root of p'(x)=0.If p(-1)<p(1),then in the interval [-1,1],which is maximum and minimum of p(-1) and p(1)?:
a)p(-1) is minimum and p(1) is maximum.
b)p(-1) is not minimum and p(1) is maximum.
c)neither p(-1) is minimum nor p(1) is maximum.
To determine the maximum and minimum values of a function in a given interval, we need to find the critical points. Critical points occur where the derivative of the function is equal to zero or undefined.
Given that x = 0 is the only real root of p'(x) = 0, it means that the derivative p'(x) has a double root at x = 0. This tells us that the derivative changes sign from negative to positive at x = 0. In other words, the function p(x) has a local minimum at x = 0.
Now, let's analyze the interval [-1, 1].
Since p'(x) has a double root at x = 0, it means p'(x) is negative (-) to the left of x = 0 and positive (+) to the right of x = 0. This implies that p(x) is decreasing from [-1, 0) and increasing from (0, 1].
Given that p(-1) < p(1), it means that p(x) is smaller at x = -1 than at x = 1. Therefore, p(-1) is the minimum value of p(x) in the interval [-1, 1], and p(1) is the maximum value of p(x) in the interval.
Hence, the correct answer is option a) p(-1) is minimum, and p(1) is maximum.