Answer to Question #156927 in Calculus for Koochika

Question

Can the intermediate value theorem can be applied to show the that there is a root of equation x^5-x^3+3x-5 in the given interval [1,2] if yes, apply it

Answer 1

The function $f(x)=x^5-x^3+3x-5$ is continuus on $\R$ as a polynomial.

Then the function $f$ is continuous on the closed interval $[1,2].$

$$f(1)=(1)^5-(1)^3+3(1)-5=-2<0$$

$$f(2)=(2)^5-(2)^3+3(2)-5=25>0$$

$$f(1)=-2<0<25<f(2)$$

Hence the Intermediate Value Theorem can be applied.

Then there exists a number $c$ in $(0, 1)$ such that $f(c)=0.$

Therefore the equation $x^5-x^3+3x-5=0$ has at least one root $c$ in the interval $(1, 2).$