hey I just wanted to make sure that I was doing this problem correctly.
The question says to graph the function f given by f(x)=x^3+3x^2-9x-13 and find the local extremer.
So here's what I did:
f(x)=x^3+3x^2-9x-13
F^1(x)=3x^2+6x-9
= 3x(x-3)
x=0
x=3
I put those two xes back into the original equation and got: (0,-13) (3,-4) and then graphed. Is this the proper way to look for local extremer?
your simplification of the derivative is not correct
F ' (x) = 3x^2 + 6x - 9
now we set that equal to zero
3x^2 + 6x-9 = 0
divide each term by 3
x^2 + 2x - 3 = 0
which factors to
(x+3)(x-1) = 0
so x = -3 or x = 1
if x=1, f(1) = 1+3-9-13 = -18
if x=-3, f(-3) = -27 + 27 + 27 - 13 = 14
so the points are (1,-18) and (-3,14)
(I don't understand how you went from
3x^2+6x-9 to
3x(x-3)
if you expand 3x(x-3) you get 3x^2 - 9x , I don't see the 6x )
You should always check intermediate steps of your solution )
yes, you have the max or min. These are the localized extremes. Look at x=+- inf? what happens there? (But they are not local extremes. Read this example to see the difference between local extrema and global extrema.
http://en.wikipedia.org/wiki/Maxima_and_minima
I took out the 3x which gave me 3x(x+6-9) but if that's wrong...it's wrong :) thanks for the help!
Andddd I just realized to take out the 3 out the 6 and 9 T_T opps. Thanks, now I feel dumb XD
Yes, you're on the right track! To find the local extrema of a function, you first need to find the critical points. Critical points occur where the derivative of the function is equal to zero or undefined. In your case, you found the derivative correctly:
f'(x) = 3x^2 + 6x - 9
Now, to find the critical points, set the derivative equal to zero and solve for x:
3x^2 + 6x - 9 = 0
You can simplify this equation further by dividing through by 3:
x^2 + 2x - 3 = 0
Now, factor the quadratic equation:
(x + 3)(x - 1) = 0
So the critical points are x = -3 and x = 1.
Next, you need to check the values of the function at these critical points and the endpoints of the interval you are considering. In this case, since the question did not provide a specific interval, you can consider the entire x-axis.
Evaluate the function at the critical points and the endpoints:
f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) - 13 = -1
f(1) = 1^3 + 3(1)^2 - 9(1) - 13 = -18
f(-∞) = as x approaches negative infinity, the function approaches negative infinity
f(+∞) = as x approaches positive infinity, the function approaches positive infinity
Now, plot the points (-3, -1) and (1, -18) on the graph and observe whether they represent local extrema or not. In this case, you have a local minimum at (-3, -1) and there is no local maximum since the function approaches positive infinity as x approaches positive infinity.
So, by analyzing the critical points and evaluating the function at those points and the endpoints, you correctly determined that (-3, -1) is the local minimum of the function. Well done!