increasing/decreasing derivatives and bounds
- Thread starter cosh
- Start date
Part c, plug values into your derivative. If it's > 0, then it's increasing. If it's negative, it's decreasing.
Part d, if f''(x)>0 on an interval, then concave up.
If f''(x)<0 on an interval, then concave down.
To find inflection points, set the 2nd derivative equal to 0 and solve for x.
Since this is a nice continuous curve, we can:
\(\displaystyle \L\\y''=(9x^{2}-12x+2)e^{-3x}\)
Note the quadratic.
\(\displaystyle \L\\9x^{2}-12x+2=0\)
Solve to find your inflection points. That's where it changes concavity.
Check points on either side of the inflection points to see whether it's up or down.
i.e. Check f''(1)=-.049787... It's concave down at x=1.
It helps to graph these to see where they're increading, decreasing, up, down, etc.
Part d, if f''(x)>0 on an interval, then concave up.
If f''(x)<0 on an interval, then concave down.
To find inflection points, set the 2nd derivative equal to 0 and solve for x.
Since this is a nice continuous curve, we can:
\(\displaystyle \L\\y''=(9x^{2}-12x+2)e^{-3x}\)
Note the quadratic.
\(\displaystyle \L\\9x^{2}-12x+2=0\)
Solve to find your inflection points. That's where it changes concavity.
Check points on either side of the inflection points to see whether it's up or down.
i.e. Check f''(1)=-.049787... It's concave down at x=1.
It helps to graph these to see where they're increading, decreasing, up, down, etc.
You have the derivative wrong. I posted the correct one.
You have the first derivative right.
\(\displaystyle \L\\f'(x)=2xe^{-3x}-3x^{2}e^{-3x}\)
Take the derivative of \(\displaystyle \L\\2xe^{-3x}\) first.
You get \(\displaystyle \L\\2e^{-3x}-6xe^{-3x}\)
Now, take derivative of the other half, you get:
\(\displaystyle \L\\6xe^{-3x}-9x^{2}e^{-3x}\)
So, you have:
\(\displaystyle \L\\f''(x)=2e^{-3x}-6xe^{-3x}-6xe^{-3x}+9x^{2}e^{-3x}\)
Factor out e^(-3x) and you get:
\(\displaystyle \L\\f''(x)=e^{-3x}(9x^{2}-12x+2)\)
You need only concentrate on the quadratic to find the roots and, therefore, the inflection points.
You have the first derivative right.
\(\displaystyle \L\\f'(x)=2xe^{-3x}-3x^{2}e^{-3x}\)
Take the derivative of \(\displaystyle \L\\2xe^{-3x}\) first.
You get \(\displaystyle \L\\2e^{-3x}-6xe^{-3x}\)
Now, take derivative of the other half, you get:
\(\displaystyle \L\\6xe^{-3x}-9x^{2}e^{-3x}\)
So, you have:
\(\displaystyle \L\\f''(x)=2e^{-3x}-6xe^{-3x}-6xe^{-3x}+9x^{2}e^{-3x}\)
Factor out e^(-3x) and you get:
\(\displaystyle \L\\f''(x)=e^{-3x}(9x^{2}-12x+2)\)
You need only concentrate on the quadratic to find the roots and, therefore, the inflection points.
