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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.)enter image description here

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.

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$f(1)>0$ 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.

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  • $\begingroup$ $f'(x)$ is not zero anywhere but curve takes a turn how? $\endgroup$ – Daman Oct 29 '18 at 16:31
  • $\begingroup$ @Damn1o1: who told you that ? $f'$ has two roots. $\endgroup$ – Yves Daoust Oct 29 '18 at 16:38
  • $\begingroup$ I am using definition of stationary point . $\endgroup$ – Daman Oct 29 '18 at 16:41
  • $\begingroup$ @Damn1o1: and... ? $\endgroup$ – Yves Daoust Oct 29 '18 at 16:43
  • $\begingroup$ "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" . Curve bends at this point. Isn't it stationary point ? $\endgroup$ – Daman Oct 29 '18 at 16:47

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