Question

could you help me with this problem of intermediate value theorem please .

Answer 1

so i got everything but in the end i must solve for b but im kind of lost how to solve it

Answer 2

You have to find \(b\) such that \(g(x)\) is continuous, right? Where does the IVT come into play?

Answer 3

on im sorry no that's not IVT

Answer 4

hahaha that's contunity i just have mix of topic here

Answer 5

yeah i must solve for b so g(x) becomes continuous

Answer 6

Okay, so you have the right limits and everything, so you just need to solve for \(b\) in
\[b=-\frac{b}{b+1}\]
Multiply both sides by \(b+1\):
\[b(b+1)=-b\]
Simplify:
\[b^2+b=-b~~\iff~~b^2+2b=b(b+2)=0\]

Answer 7

so be can be two values ? 0 and -2

Answer 8

b*

Answer 9

It's certainly true if \(b=0\). If that's the case, then
\[g(x)=\begin{cases}x\\x^2\end{cases}\]
for the appropriate domains since they "intersect" at the origin.

If \(b=-2\), you have
\[g(x)=\begin{cases}\dfrac{x+2}{3}\\x^2-2\end{cases}\]
for the appropriate domains. As \(x\to0\), you get \(g(x)\to\dfrac{2}{3}\) as \(x\to0^-\) and \(g(x)\to-2\) as \(x\to0^+\), which means this \(g\) can't be continuous.

Answer 10

Oops, that second \(g\) should be
\[g(x)=\begin{cases}\color{red}{-(x+2)}\\x^2-2\end{cases}\]
which WOULD be continuous as \(x\to0\) from either side. Sorry about that

Answer 11

so both value of b work right

Answer 12

Thank You Firiend -----!!