Answer
$f(0)=-1$ and $f(1)=10$
This shows that $f(0)$ and $f(1)$ attains opposite variations.
So, as per the Intermediate Value Theorem, there must be a real zero in the interval $[0,1]$.
Work Step by Step
We are given that $f(x)=8x^4-2x^2+5x-1$
The Intermediate Value Theorem states that when a function is continuous on an interval $[p,q]$ and takes values $f(p)$ and $f(q)$ at the endpoints, then the function takes all values between $f(p)$ and $f(q)$ at some point of the interval.
We will evaluate the function at the endpoints $[0,1]$.
$f(0)=8(0)^4-2(0)^2+5(0)-1=0-0+0-1=-1$
$f(1)=8(1)^4-2(1)^2+5(1)-1=8-2+5-1=10$
This shows that $f(0)$ and $f(1)$ attains opposite variations.
So, as per the Intermediate Value Theorem, there must be a real zero in the interval $[0,1]$.
