Sunday, October 4, 2009

Post #7

This week in calculus we did optimization for the first three days, had a field trip Thursday, and worked on our review packets Friday so we really did not learn anything new. I am going to review how to find the maxs and mins using the shortcut method used for multiple choice questions only because I need a review on how to do it so yall might too.

The first step is to take the derivative of the equation given. Then you set the derivative equal to zero and solve for x. After, you take the second derivative and plug in each of your x's into it. If your answer turns out to be a positive number then it means the second derivative is concave up and it is a min. If it is a negative number, it means it is concave down and it is a max. If it equals zero, it means it is neither a max nor min.

EXAMPLE:

f(x) = -3x^5 + 5x^3

f'(x) = -15x^4 + 15x^2
= -15x^2 (x^2-1)
= -15x^2 (x+1) (x-1) = 0

x= -1, 0, 1

f''(x)= -60x^3 + 30x

-1: -60 (-1)^3 + 30 (-1) = positive; concave up
0: -60 (0)^3 + 30 (0) = 0
1: -60 (1)^3 + 30 (1) = negative; concave down

Therefore, there is a min at x= -1 and a max at x=1

I am having trouble with how to find a tangent line to a graph. I get stuck after I take the derivative. I used to know how to do it, but I forgot. If anybody can help that would be great.

1 comment:

  1. remember you have to find slope. so take the derivative of the equation. Then plug in the x value from the point given to get your slope. Then use the slope and point and plug into slope intercept form which is
    y-y1=slope(x-x1)

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