samantha0417
New member
- Joined
- Sep 24, 2006
- Messages
- 12
find all critical numbers and use the second derivative test to determine if any of the CPs are relative maxima and minima
f(x)=(1/x) - [1/(x+4)]
f(x)=(1/x) - [1/(x+4)]
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
second derivative test
- Thread starter samantha0417
- Start date
Hello Samantha.
Find your first derivative. Set to 0 and solve for x.
Find the 2nd derivative. Use the x value you found in the first derivative.
Is it negative?. If so, a relative max.
Is it positive?. If so, a relative min.
Graph your function. You can see then where the concavities lie.
Find your first derivative. Set to 0 and solve for x.
Find the 2nd derivative. Use the x value you found in the first derivative.
Is it negative?. If so, a relative max.
Is it positive?. If so, a relative min.
Graph your function. You can see then where the concavities lie.
Hello, Samantha!
Galactus gave you an excellent game plan.
Here's my two cents . . .
Combine the two fractions and work with: \(\displaystyle \L\,f(x) \;= \;\frac{4}{x(x\,+\,4)}\)
Better yet: \(\displaystyle \L\,f(x) \;= \;\frac{4}{x^2\,+\,4x} \;= \;4\left(x^2\,+\,4x\right)^{-1}\)
Galactus gave you an excellent game plan.
Here's my two cents . . .
Find all critical numbers and use the second derivative test
to determine if any of the CPs are relative maxima and minima.
. . . \(\displaystyle \L f(x)\;=\;\frac{1}{x}\,-\,\frac{1}{x+4}\)
Combine the two fractions and work with: \(\displaystyle \L\,f(x) \;= \;\frac{4}{x(x\,+\,4)}\)
Better yet: \(\displaystyle \L\,f(x) \;= \;\frac{4}{x^2\,+\,4x} \;= \;4\left(x^2\,+\,4x\right)^{-1}\)