Question

Let $f\left(x\right)=(x-4)(x-5)(x-6)(x-7)$ then

• A. $f\left(x\right)=0$ has four roots
• B. three roots of $f^{\prime }\left(x\right)=0$ lie in $\left(4,5\right)\cup \left(5,6\right)\cup \left(6,7\right)$
• C. The equation $f^{\prime }\left(x\right)=0$ has only one root
• D. three roots of $f^{\prime }\left(x\right)=0$ lie in $\left(3,4\right)\cup \left(4,5\right)\cup \left(5,6\right)$

Answer 1

The correct option is A $f\left(x\right)=0$ has four roots

$f\left(x\right)=\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)$

then $f\left(x\right)=0$

$\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=0$

$\Rightarrow$ $x=4,5,6,7$

Has 4 root.