# Concavity

• Nov 15th 2009, 09:15 AM
ctran
Concavity
For the function f(x)=(x^(2)+4x+(11/4)e^(x) I need to find where it is concave up.

So, I found the first derivative and then the second derivative which is,
f''(x)= e^(x)(x^(2)+8x+(51/4))
Then, I found the roots of the function using the quadratic formula which is (-8+/-sqrt(13))/(2)

I found it was concave upwards at the open interval: (((-8+sqrt(13))/2), inf)

But somehow this is incorrect. Can someone help me?
• Nov 15th 2009, 09:24 AM
skeeter
Quote:

Originally Posted by ctran
For the function f(x)=(x^(2)+4x+(11/4)e^(x) I need to find where it is concave up.

So, I found the first derivative and then the second derivative which is,
f''(x)= e^(x)(x^(2)+8x+(51/4))
Then, I found the roots of the function using the quadratic formula which is (-8+/-sqrt(13))/(2)

I found it was concave upwards at the open interval: (((-8+sqrt(13))/2), inf)

But somehow this is incorrect. Can someone help me?

is the function ...

$\displaystyle f(x) = \left(x^2 + 4x + \frac{11}{4}\right)e^x$

or

$\displaystyle f(x) = x^2 + 4x + \frac{11}{4}e^x$
• Nov 15th 2009, 09:28 AM
ctran
Quote:

Originally Posted by skeeter
is the function ...

$\displaystyle f(x) = \left(x^2 + 4x + \frac{11}{4}\right)e^x$

or

$\displaystyle f(x) = x^2 + 4x + \frac{11}{4}e^x$

Hello skeeter,
The function is this one,

$\displaystyle f(x) = \left(x^2 + 4x + \frac{11}{4}\right)e^x$
• Nov 15th 2009, 09:32 AM
skeeter
Quote:

Originally Posted by ctran
Hello skeeter,
The function is this one,

$\displaystyle f(x) = \left(x^2 + 4x + \frac{11}{4}\right)e^x$

your 2nd derivative is incorrect ...

$\displaystyle f''(x) = e^x\left(x^2+6x+\frac{27}{4}\right)$
• Nov 15th 2009, 09:37 AM
ctran
Quote:

Originally Posted by skeeter
your 2nd derivative is incorrect ...

$\displaystyle f''(x) = e^x\left(x^2+6x+\frac{27}{4}\right)$

No I dont think so, $\displaystyle f''(x) = e^x\left(x^2+6x+\frac{27}{4}\right)$ is actually the first derivative.

I am certain the second derivative is $\displaystyle f''(x) = e^x\left(x^2+8x+\frac{51}{4}\right)$
• Nov 15th 2009, 09:48 AM
skeeter
Quote:

Originally Posted by ctran
No I dont think so, $\displaystyle f''(x) = e^x\left(x^2+6x+\frac{27}{4}\right)$ is actually the first derivative.

I am certain the second derivative is $\displaystyle f''(x) = e^x\left(x^2+8x+\frac{51}{4}\right)$

you're right ... I copied the wrong one down.

$\displaystyle f''(x) = e^x\left(x^2+8x+\frac{51}{4}\right)$

$\displaystyle f''(x) = 0$ at $\displaystyle x = \frac{-8 \pm \sqrt{13}}{2} \approx -2.2$ and $\displaystyle -5.8$

for x < -5.8 , f''(x) > 0 ... concave up

for -5.8 < x < -2.8 , f''(x) < 0 ... concave down

for x > -2.8 , f''(x) > 0 ... concave up