Answer to Question #227181 in Calculus for nox

Question

Find an interval where the following functions has a root. Show all you work how you chose the interval. i) f(x) = 3x + sin(x) − e^x

Answer 1

Use the Intermediate Value Theorem

The function $f(x)=3x+\sin(x)-e^x$ is continuous on $(-\infin, \infin).$ Then the function $f$ is continuous on $[0,1].$

$$f(0)=3(0)+\sin(0)-e^0=-1<0$$

$$f(1)=3(1)+\sin(1)-e^1>0$$

By the Intermediate Value Theorem $\exist c\in(0,1)$ such that $f(c)=0.$

The function $f(x)=3x+\sin(x)-e^x$ is continuous on $[1, 2].$

$$f(2)=3(2)+\sin(2)-e^2<7-e^2<0$$

By the Intermediate Value Theorem $\exist c\in(1,2)$ such that $f(c)=0.$

$$f'(x)=3+\cos(x)-e^x$$

$$f'(x)=3+\cos(x)-e^x>3+\cos(x)-1>0, x\leq0$$

$$f'(x)=3+\cos(x)-e^x>3+\cos(x)-e>0, 0\leq x\leq1$$

Hence $f(x)$ strictly increases on $(-\infin, 1)$

Therefore the function $f$ has the only root on $(-\infin, 1].$

Therefore we choose the interval $[0, 1]$ to find the root of the function $f.$

$$f'(x)=3+\cos(x)-e^x\leq4-e^2<0, x\geq2$$

Hence $f(x)$ strictly decreases on $(2, \infin)$

Therefore the function $f$ has no root on $[2, \infin).$

Therefore we choose the interval $[1, 2]$ to find the root of the function $f.$

Therefore we choose the interval $[0, 2]$ to find all roots of the function $f.$