calculus question?

Question

find the max and min values of this funtion by taking the derivaive and set it equal to zero plz tell me in detail .i got in stuck in some step. f(x)=-1/8 (x+5) (x+4) (x+3) (x+2)

Answer 1

f(x) = (-1/8) (x + 5) (x + 4) (x + 3) (x + 2)

Instead of expanding out this mess of a problem, what I'm going to do is multiply out the first two and the last two, retaining the brackets.

f(x) = (-1/8) (x^2 + 9x + 20) (x^2 + 5x + 6)

In order to find the min/max values, I'm going to take the derivative of this modified (but still the same) function. I'm going to use the product rule while ignoring (-1/8) {since it is a constant and we can do this when taking the derivative}.

f'(x) = (-1/8) { [2x + 9](x^2 + 5x + 6) + (x^2 + 9x + 20)(2x + 5) }

Expanding the stuff in the brackets,

f'(x) = (-1/8) [2x^3 + 10x^2 + 12x + 9x^2 + 45x + 54 + 2x^3 + 5x^2 + 18x^2 + 45x + 40x + 100]

f'(x) = (-1/8) [4x^3 + 42x^2 + 142x + 154]

Our cubic has all even terms. We can factor a 2 out.

f'(x) = (-1/8)(2) [2x^3 + 21x^2 + 71x + 72]
f'(x) = (-1/4) [2x^3 + 21x^2 + 71x + 72]

Solving for the roots will be a difficult process... in progress.
Now that it has been revealed the roots aren't easy to solve for, I'll stop at this point.

Answer 2

Hi, i will try to answer your quesiton with detail.

First you must know that derivating an equation, gives you the slope of any point of the given equation, so this means that, if the slope is "0", the graph of the given equation at that point represents a minimum, or a maximum, since there is no more rising, nor decreasing.

So to find a maximum and a minimum you first must have tu derivate the equation and then equal it to zero, and solve for X, then plug those values of X in the original equaiton to solve for Y.

f(x)= -1/8 (x+5) (x+4) (x+3) (x+2)

To simplify the derivation, first we simplify the equation

f(x)= -1/8 (x^2+9x+20)(x^2+5x+6)

f(x)= -1/8 (x^4+5x^3+6x^2+9x^3+45x^2+54x+20x^2+100x+120)

Then we add like-terms

f(x)= -1/8 (x^4+14x^3+71x^2+154x+120)

Now we derivate:

f '(x)= -1/8 (4x^3+42x^2+142x+154)

Finally we equal to 0 and solve for X's values(to find where the slope equals 0)

0 = -1/8 (4x^3+42x^2+142x+154)

0 = 4x^3+42x^2+142x+154

I don't have a calculator, but the X values should be arround, -2.8 and -4.1.

If you substitute these values in the original you get..

f(-2.8)= -1/8 (-2.8+5) (-2.8+4) (-2.8+3) (-2.8+2)= -2.1

and

f(-4.1)= -1/8 (-4.1+5) (-4.1+4) (-4.1+3) (-4.1+2)= 2.1

So the points are (-2.8,-2.1) and (-4.1,2.1) which makes (-4.1,2.1) a maximum and (-2.8,-2.1) a minimum.

Hope that helps =)

Answer 3

hmmm
f(x) = -1/8 (x^2 + 9x + 20) (x^2 + 5x + 6)
= -1/8 (x^4 + 5x^3 +6x^2 + 9x^3 + 45x^2 + 54x + 20x^2 + 100x + 120)
= -1/8 (x^4 + 14x^3 + 71x^2 + 154x + 120)
f'(x) = -1/8 (4x^3 + 42x^2 + 142x + 154) = 0
2x^3 + 21x^2 + 71x + 77 = 0
graphing calc solve for x
x = -4.618
f''(x) = -1/8 (12x^2 + 84x + 142)
f''(-4.618) = -1.25, meaning concave down, so -4.618 is the local maximum
=)

Answer 4

I took calculus...it ain't worth taking...no job requires it. You + Me = calculus.

Answer 5

yes somone help me help him