# To find number of roots in a given interval of a cubic equation.

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.

$$f(1)>0$$ 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.
• $f'(x)$ is not zero anywhere but curve takes a turn how? – Daman Oct 29 '18 at 16:31
• @Damn1o1: who told you that ? $f'$ has two roots. – Yves Daoust Oct 29 '18 at 16:38
• "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 ? – Daman Oct 29 '18 at 16:47