Solve using intermediate value theorem: Show that x^3-5 x^2+3 x+6…

Question

Solve using intermediate value theorem:
Show that $x^3-5 x^2+3 x+6=0$ has at least two real root between $x=1$ and $x=5$

✓

Answer

Let $f(x)=x^3-5 x^2+3 x+6$ which is a polynomial function and hence continuous on $[1,5]$
We factorise $f(x)$ by synthetic division:
$ \therefore f(x)=(x-2)\left(x^2-3 x-3\right)$
$\therefore f(1)=(1-2)(1-3-3)=5>0$
$f(2)=(2-2)(4-6-3)=0$
$\therefore x =2 \text { is a root of } f ( x )=0$
$f(3)=(3-2)(9-9-3)=-3<0$
$f(4)=(4-2)(16-12-3)=2>0$
$f$ is continuous on $[3,4]$
$f(3)<0, f(4)>0$
$\therefore$ by intermediate value theorem for continuous function $f(x)=0$ has a root between $3$ and $4$
$\therefore$ there are two roots, $x=2$ and a root between $x=3$ and $x=4$.
$\therefore f(x)=0$ has at least two root between $1$ and $5.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Solve the Following Question.(3 Marks)" from Continuity→Continuity — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Continuity — Maths STD 11 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions