Question

suppose you wish to show that the equation $xcos(x)=ln(x)$ has at least one solution. equivalently, we can show that $xcos(x)-ln(x)=0$ has a solution, so we define $f(x)=xcos(x)-ln(x)$. it is enough to show that $f(x)$ crosses the line $y = 0$. (a) drag the two sliders to choose an interval $a,b$ where it appears that $f(a)$ and $f(b)$ are on opposite sides of the line $y = 0$. you have chosen the interval $1.0,5.0$. (b) evaluate $f(x)$ at both endpoints of the interval. $f(1.0)=0.54$ $f(5.0)= - 0.191$ (c) what property of $f(x)$ ensures that the graph crosses $y = 0$ on this interval? $f(x)$ is continuous on the interval $1.0,5.0$. (d) state your conclusion. by the intermediate - value theorem, since $f(x)$ is continuous on $1.0,5.0$ and $f(5.0)<0

Explanation:

Step1: Evaluate $f(1.0)$

We know that $f(x)=x\cos(x)-\ln(x)$. Substitute $x = 1.0$ into the function: $f(1.0)=1.0\times\cos(1.0)-\ln(1.0)$. Since $\cos(1.0)\approx0.54$ and $\ln(1.0) = 0$, then $f(1.0)=1\times0.54 - 0=0.54$.

Step2: Evaluate $f(5.0)$

Substitute $x = 5.0$ into the function $f(x)=x\cos(x)-\ln(x)$. We have $\cos(5.0)\approx0.284$ and $\ln(5.0)\approx1.609$. Then $f(5.0)=5.0\times0.284-1.609=1.42 - 1.609=- 0.189\approx - 0.191$.

Step3: Apply Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function $y = f(x)$ 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 the open interval $(a,b)$ such that $f(c)=k$. Here, $a = 1.0$, $b = 5.0$, $f(1.0)=0.54>0$, $f(5.0)=-0.191<0$, and $k = 0$. Since $f(x)$ is continuous on $[1.0,5.0]$, there exists a number $c$ in $(1.0,5.0)$ such that $f(c)=0$.

Answer:

Since $f(x)$ is continuous on $[1.0,5.0]$, $f(1.0)=0.54>0$, $f(5.0)=-0.191<0$, by the Intermediate - Value Theorem, there exists a number $c$ in $(1.0,5.0)$ such that $f(c) = 0$.