Say if the following equations have a solution using Bolzano’s theorem:

a) $$x^2=1$$

b) $$e^x= 3+\ln x$$

c) $$x^4+2x=0$$

### Development:

a) We define the function $$f(x)=x^2-1$$. We are going to look for two values $$a$$ and $$b$$ such that once we evaluate the function $$f (x)$$ we obtain values with opposite signs:

Taking

$$x=0 \Rightarrow f(0)=-1 < 0$$

$$x=2 \Rightarrow f(2)=5 > 0$$

so in the interval $$[0,2]$$ a point $$c$$ exists such that $$f (c) = 0$$ and therefore $$c$$ is a solution to our equation. (in this case $$c=1$$ and $$f (1) = 0$$).

b) We define the function $$f(x)=e^x-\ln x-3$$. Let's look for two values $$a$$ and $$b$$ such that once we evaluate the function $$f (x)$$ we obtain values with opposite signs:

Taking

$$x=1 \Rightarrow f(1)=e-0-3=-0.2817 < 0$$

$$x=2 \Rightarrow f(2)=3.69 > 0$$

So in the interval $$[1,2]$$ a point $$c$$ exists where $$f (c) = 0$$ and we know with certainty that some value that solves our equation exists.

c) We define the function $$f(x)=x^4+2x$$ and repeat the process:

Taking

$$x=-1 \Rightarrow f(-1)=(-1)^4+2\cdot(-1)=1-2=-1 < 0$$

$$x=1 \Rightarrow f(1)=1+2=3 > 0$$

so in the interval $$[-1,1]$$ there exists a point $$c$$ that is a solution to our equation.

### Solution:

a) It has at least one solution in the interval $$[0,2]$$.

b) It has at least one solution in the interval $$[1,2]$$.

c) It has at least one solution in the interval $$[-1,1]$$.