Local extremes/concavity

[MarkSA]

Here are two final problems:

1) For what values of a and b does the function:
f(x) = x^3 + ax^2 + bx + 2
have a local maximum when x = -3 and a local minimum when x = -1?

I'm not sure how to solve this one. So far I had:
"To have a local max/min, f'(x) must be a crit value (0 or undefined), since f(x) is a polynomial 0 is the only possibly case. So f'(-3) and f('-1) must be zero. Derivative of f'(x) is 3x^2 + (a)2x + b. So if I substitute -3 and -1 for x I can say:
[3(-3)^2 + (a)2(-3) + b] + [3(-1)^2 + (a)2(-1) + b] = 0
or
4a - b = 15
But i'm still stuck on how to get a and b. It would be easier to guess at now with a calculator (maybe), but there must be another way.

2) (Assume all functions are twice differentiable and the 2nd derivatives are never 0)
Then, if f and g are concave upward on I, show that f + g is concave upward on I.

If f and g are concave up, then f'' and g'' must be >0 by the concavity test. But how can I use that to show that the addition of functions f + g is concave up too?

 

[galactus]

Use your info to create two equations and solve for a and b.

The derivative of said cubic is

\(\displaystyle \L\\3x^{2}+2ax+b\)

So, use your info:

\(\displaystyle \L\\3(-3)^{2}+2a(-3)+b=0\)
\(\displaystyle \L\\3(-1)^{2}+2a(-1)+b=0\)

Now, solve for a and b.

 

[MarkSA]

Yes! I just solved that one finally. Instead of setting f'(-3) + f'(-1) = 0 I did f'(-3) = f'(-1) with the appropriate x values. This cancelled out the b's and left me with the right answer for a. Then I just plugged it in and solved for b and the calculator agrees (a = 6, b = 9)

I've still gotten no insight as to how I can do the second problem though.

 

[Deleted member 4993]

2) (Assume all functions are twice differentiable and the 2nd derivatives are never 0)
Then, if f and g are concave upward on I, show that f + g is concave upward on I.

If f and g are concave up, then f'' and g'' must be >0 by the concavity test. But how can I use that to show that the addition of functions f + g is concave up too?

(f+g)" = f" + g"

 

[galactus]

2) (Assume all functions are twice differentiable and the 2nd derivatives are never 0)
Then, if f and g are concave upward on I, show that f + g is concave upward on I.

If f and g are concave up, then f'' and g'' must be >0 by the concavity test. But how can I use that to show that the addition of functions f + g is concave up too?[/quote]


If \(\displaystyle \L\\x_{1}<x_{2}\), where \(\displaystyle \L\\x_{1} \;\ and \;\ x_{2}\) are in I, then \(\displaystyle \L\\f(x_{1})<f(x_{2}) \;\ and \;\ g(x_{1})<g(x_{2})\\\), therefore, \(\displaystyle \L\\f(x_{1})+g(x_{1})<f(x_{2})+g(x_{2}), \;\ (f+g)(x_{1})<(f+g)(x_{2}).\)

Therefore, hence, whence and so, f+g is increasing on I.

 

[daon]

If f, g are concave up then f'' > 0 and g'' > 0 as was stated.

Note that (f+g)'' = ((f+g)')' = (f'+g')' = f'' + g'' > 0 + 0 = 0.