Question

1. use the intermediate value theorem to show that the equation f(x) = 0 has at least one real solution where f(x) = cos(πx) + ln(x). see examples 1.9.1 and 1.9.2 in chads calculus notes

Explanation:

Step1: Determine the domain of \(f(x)\)

The domain of \(y = \ln(x)\) is \((0,+\infty)\), so the domain of \(f(x)=\cos(\pi x)+\ln(x)\) is \((0,+\infty)\).

Step2: Evaluate \(f(x)\) at two - points

Let's choose \(x = 1\) and \(x=\frac{1}{2}\).
When \(x = 1\), \(f(1)=\cos(\pi\times1)+\ln(1)\). Since \(\cos(\pi)=- 1\) and \(\ln(1) = 0\), then \(f(1)=-1+0=-1<0\).
When \(x=\frac{1}{2}\), \(f(\frac{1}{2})=\cos(\pi\times\frac{1}{2})+\ln(\frac{1}{2})\). Since \(\cos(\frac{\pi}{2}) = 0\) and \(\ln(\frac{1}{2})=-\ln(2)\approx - 0.693<0\). Let's choose another point, say \(x = e\). Then \(f(e)=\cos(\pi e)+\ln(e)\). Since \(\ln(e) = 1\) and \(-1\leqslant\cos(\pi e)\leqslant1\), \(f(e)=\cos(\pi e)+1\geqslant0\). If \(\cos(\pi e)= - 1\), \(f(e)=0\). If \(\cos(\pi e)>-1\), \(f(e)>0\).

Step3: Apply the Intermediate - Value Theorem

The function \(y = f(x)=\cos(\pi x)+\ln(x)\) is continuous on the interval \((1,e)\) (because \(y = \cos(\pi x)\) and \(y=\ln(x)\) are both continuous on \((1,e)\)). Since \(f(1)<0\) and \(f(e)\geqslant0\), by the Intermediate - Value Theorem, which states that if \(y = f(x)\) is continuous on a closed interval \([a,b]\) and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\in(a,b)\) such that \(f(c)=k\). In the case of \(k = 0\), since \(0\) is between \(f(1)\) and \(f(e)\), there exists at least one \(c\in(1,e)\) such that \(f(c)=0\).

Answer:

The equation \(f(x)=\cos(\pi x)+\ln(x)=0\) has at least one real - solution because the function \(f(x)\) is continuous on an appropriate interval (e.g., \((1,e)\)) and \(f(x)\) takes on values of opposite signs at the endpoints of that interval, so by the Intermediate - Value Theorem, there is a point \(c\) in the interval where \(f(c)=0\).