The equation has $f(x)=x^3 +9x^2-49x+49$
$1$. exactly one root in the interval $(1,2)$
$2$. exactly two distinct roots in the interval $(1,2)$
$3$. exactly three distinct roots in the interval $(1,2)$
$ 4$. No roots in the interval $(1,2)$
My input
$f'(x)=3x^2+18x-49=0$
Discriminant=$D$ of this equation: $D<0$ $\implies $ imaginary roots meaning that no stationary points(I have one doubt here: doubt is that when I plotted the graph of it in desmos there is one stationary point at $(2.033,-5.017)$ as shown below I don't understand why because we are not getting any value of $x$ for which our derivative is zero.)
Moving further. $f(1)=10$ and $f(2)=-5$ meaning that graph must cross $x$-axis between $(1,2)$ at least once but we don't get any stationary point from our derivative that makes it only one root between $(1,2)$ so option $1$.
I found this method somewhat sloppy and time-consuming.This question came as one mark in my objective exam. Please, someone, suggest me a proper way to solve this kind of problems.
Best Answer
$f(1)>0$ and $f(2)<0$ indeed indicate $1$ or $3$ roots in the given interval. Then $f'(1)= -28$ and $f'(2)=-1$ show that the extrema are outside $[1,2]$, as the parabola goes up. So the answer is one.