Answer to Question #295391 in Calculus for Fresh

Question

Show that x^5 +4x=1 has exactly one solution on [0,1]

Answer 1

$$f(x)=x^5 +4x-1$$

The function $f(x)$ is continuous on $\R$ as polynomial.

$$f(0)=0+0-1=-1<0$$

$$f(1)=1+4-1=4>0$$

By the Intermediate Value Theorem there exists a number $c\in (0, 1)$ such that $f(c)=0.$

$$f'(x)=5x^4+4,$$

Since $f'(x)>0$ for $x\ge0,$ the function is strictly increasing on $[0, \infin).$ Hence there is the only number $c\in (0, 1)$ such that $f(c)=0.$

Therefore the equation $x^5+4x=1$ has exactly one solution on $[0,1].$