Semaj Christian

2022-06-21

Using the Intermediate Value Theorem and a calculator, find an interval of length 0.01 that contains a root of ${x}^{5}-{x}^{2}+2x+3=0$, rounding off interval endpoints to the nearest hundredth.

I've done a few things like entering values into the given equation until I get two values who are 0.01 apart and results are negative and positive ($-1.15$ & $-1.16$), but these answers were incorrect.

I'm at the point where I'm thinking there is not enough information to solve. Any ideas?

I've done a few things like entering values into the given equation until I get two values who are 0.01 apart and results are negative and positive ($-1.15$ & $-1.16$), but these answers were incorrect.

I'm at the point where I'm thinking there is not enough information to solve. Any ideas?

humbast2

Beginner2022-06-22Added 21 answers

$f(x)={x}^{5}-{x}^{2}+2x+3$

As you can see $f(0)=3>0$ and $f(-1)=-1<0$

Thus there is at least one root of $f(x)=0$ in Interval $(-1,0)$

Now calculate the value of

$f(-\frac{1}{2})=\frac{55}{32}>0$

Thus now our interval is shortened and it is $(-1,-\frac{1}{2})$

$f(-\frac{3}{4})=\frac{717}{1024}>0$

Our interval is now $(-1,-\frac{3}{4})$

$f(-\frac{7}{8})=-\frac{935}{32768}<0$

Our interval is now $(-\frac{7}{8},-\frac{3}{4})$

similarly, keep doing until you get the desired result

As you can see $f(0)=3>0$ and $f(-1)=-1<0$

Thus there is at least one root of $f(x)=0$ in Interval $(-1,0)$

Now calculate the value of

$f(-\frac{1}{2})=\frac{55}{32}>0$

Thus now our interval is shortened and it is $(-1,-\frac{1}{2})$

$f(-\frac{3}{4})=\frac{717}{1024}>0$

Our interval is now $(-1,-\frac{3}{4})$

$f(-\frac{7}{8})=-\frac{935}{32768}<0$

Our interval is now $(-\frac{7}{8},-\frac{3}{4})$

similarly, keep doing until you get the desired result