Question

● ex 1. use the ivt to show that the equation ( x-cos x = 0 ) has a solution between ( x = 0 ) and ( x=\frac{pi}{2} ).

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

Explanation:

Step1: Define the function for the first equation

Let \(f(x)=x - \cos x\).

Step2: Evaluate the function at the endpoints

• When \(x = 0\), \(f(0)=0-\cos(0)=0 - 1=- 1\).
• When \(x=\frac{\pi}{2}\), \(f(\frac{\pi}{2})=\frac{\pi}{2}-\cos(\frac{\pi}{2})=\frac{\pi}{2}-0=\frac{\pi}{2}\).

Step3: Apply the Intermediate - Value Theorem (IVT)

Since \(y = f(x)\) is continuous (as \(y=x\) and \(y = \cos x\) are continuous functions and the difference of two continuous functions is continuous) on the interval \([0,\frac{\pi}{2}]\), and \(f(0)=-1<0\) and \(f(\frac{\pi}{2})=\frac{\pi}{2}>0\). By the Intermediate - Value Theorem, there exists a number \(c\in(0,\frac{\pi}{2})\) such that \(f(c) = 0\), i.e., \(c-\cos c=0\).

Step4: Define the function for the second equation

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

Step5: Evaluate the function at some points

• When \(x=-2\), \(g(-2)=(-2)^{5}+4(-2)^{3}-7(-2)+14=-32-32 + 14+14=-36\).
• When \(x=-1\), \(g(-1)=(-1)^{5}+4(-1)^{3}-7(-1)+14=-1-4 + 7+14=16\).

Step6: Apply the Intermediate - Value Theorem (IVT)

Since \(y = g(x)\) is a polynomial function, and polynomial functions are continuous everywhere (on \((-\infty,\infty)\)). Also, \(g(-2)=-36<0\) and \(g(-1)=16>0\). By the Intermediate - Value Theorem, there exists a number \(d\in(-2,-1)\) such that \(g(d)=0\). So the equation \(x^{5}+4x^{3}-7x + 14 = 0\) has at least one real solution.

Answer:

1. By the Intermediate - Value Theorem, the equation \(x-\cos x = 0\) has a solution in \((0,\frac{\pi}{2})\).
2. By the Intermediate - Value Theorem, the equation \(x^{5}+4x^{3}-7x + 14 = 0\) has at least one real solution.