Question

Find any critical numbers of the function g(t)=t((sqrt)(2-t)), t 11 years ago

Answer 1

Any one? I cannot find anyone who knows this material so I was hoping someone on here would know. :(

Answer 2

\[g(t)=t\sqrt{2-t}~~?\]

Answer 3

yes and t<2

Answer 4

Okay, what's your derivative?

Answer 5

-1/2t(2-t)^-1/2+(2-t)^1/2

Answer 6

Okay, so your critical values are the \(t\) that make the derivative zero or undefined.

You don't have to worry about the undefined part. The condition that \(t<2\) guarantees that you have a continuous function.

Solve for \(t\):
\[-\frac{1}{2}t(2-t)^{-1/2}+(2-t)^{1/2}=0\]
Factor out a power or \((2-t)^{-1/2}\):
\[(2-t)^{-1/2}\left(-\frac{1}{2}t(2-t)^{0/2}+(2-t)^{2/2}\right)=0\]
\[(2-t)^{-1/2}\left(-\frac{1}{2}t+2-t\right)=0\]

Answer 7

how do you find the critical numbers from that? I dont know how to simplify the (2-t)^-1/2

Answer 8

Use the zero product property. If \(ab=0\), then either \(a=0\) or \(b=0\).

To apply this here, it's the same as separating your factors and setting up equations where both factors equal zero:
\[\begin{cases}
\dfrac{1}{\sqrt{2-t}}=0\\\\
-\dfrac{3}{2}t+2=0\end{cases}\]
The first equation is obtained from some minor rewriting, \((2-t)^{-1/2}=\dfrac{1}{(2-t)^{1/2}}=\dfrac{1}{\sqrt{2-t}}\).

As far as the first equation is concerned, there are no critical values. No value of \(\t\) will yield 0 for \(\dfrac{1}{\sqrt{2-t}}\), so you can ignore that equation.

Answer 9

Oh okay thank you so much!