Question

1. show that the equation ( x^{5}+4x^{3}-7x + 14 = 0 ) has at least one real solution.

Explanation:

Step1: Define the function

Let \(f(x)=x^{5}+4x^{3}-7x + 14\).

Step2: Check the values of the function at two points

• When \(x=-2\):

\(f(-2)=(-2)^{5}+4\times(-2)^{3}-7\times(-2)+14\)
\(=-32-32 + 14+14\)
\(=-36\)

• When \(x=-1\):

\(f(-1)=(-1)^{5}+4\times(-1)^{3}-7\times(-1)+14\)
\(=-1-4 + 7+14\)
\(=16\)

Step3: Apply the Intermediate Value Theorem

Since \(f(x)\) is a polynomial, it is continuous on the interval \([-2,-1]\). And \(f(-2)=-36<0\) and \(f(-1)=16>0\). By the Intermediate Value Theorem, there exists a number \(c\in(-2,-1)\) such that \(f(c) = 0\).

Answer:

The equation \(x^{5}+4x^{3}-7x + 14 = 0\) has at least one real solution.