Question

Let $f\left(x\right)=8x^3-6x^2-2x+1$, then

• A. $f\left(x\right)=0$ has no root in $(0,1)$
• B. $f\left(x\right)=0$ has at least one root in $(0,1)$
• C. $f^{\prime }\left(c\right)$ vanishes for some $c$ $ϵ$ $(0,1)$
• D. None

Answer 1

Answer: (B), (C)

$f\left(x\right)=0$ has at least one root in $(0,1)$
$f^{\prime }\left(c\right)$ vanishes for some $c$ $ϵ$ $(0,1)$

$f\left(x\right)=8x^3-6x^2-2x+1$

Now

$f\left(0\right)=1$ and
$f\left(1\right)=1$

Hence

$f\left(0\right)=f\left(1\right)$.

Applying Rolle's theorem in the interval of $[0,1]$, we get
$f^{\prime }\left(c\right)=0$ where $cϵ[0,1]$.