samantha0417
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- Sep 24, 2006
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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)]
second derivative test
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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: \(\displaystyle \L\,f(x) \;= \;\frac{4}{x^2\,+\,4x} \;= \;4\left(x^2\,+\,4x\right)^{-1}\)