Question

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Answer 1

what class is this for?

Answer 2

calc

Answer 3

so what's f'(x)?

Answer 4

x^4/4?

Answer 5

What's derivative of the function?

Answer 6

ie, if
\[g(x) = x^2 \quad ; \quad g'(x)=\frac{\mathrm{d} g}{\mathrm{d} x} = 2x\]
does that look familiar?

Answer 7

yeah but i missed the lecture so im somewhat lost ;/ can you walk me through it?

Answer 8

When you take the derivative of a normal polynomial function, the rules are pretty simple. All you're doing for powers greater than or less than 0 is reducing the power of the variable by 1, and multiplying that variable by the original power - any constants go to zero..
\[f(x)=x^n+C \longrightarrow f'(x)=\frac{\mathrm{d}f}{\mathrm{d}x} = nx^{n-1}\]

I.e.
\[g(x)=-x^2 \quad ; \quad g'(x)=\frac{\mathrm{d}g}{\mathrm{d}x}=-2x^{2-1}=-2x\]

\[h(x)=2x^5-2x \quad ; \quad h'(x)=\frac{\mathrm{d}h}{\mathrm{d}x}=5(2x^{5-1})-1(2x^{1-1})=10x^4-2\]
\[p(x) = 5x+2 \quad ; \quad p'(x)=\frac{\mathrm{d}p}{\mathrm{d}x}=1(5x^{1-1}+0)=5\]


Taking the derivative of a function gives you and equation for the line tangent to the curve of the original function at all points along the line. So, looking at
\[g(x) = -x^2\]
This is what the original function looks like...

Answer 9

so the max is 0 based off that then?

Answer 10

Sorry sorry, the derivative gives you the *slope* of the tangent line.

But yeah, having the derivative equal zero denotes either a maximum or a minimum; in the case of this parabola, it's a maximum.

Answer 11

because the slope of the line tangent to a maximum or minimum will by definition be zero - it's specifically where the slope of the curve changes from a positive to a negative, or vice versa.

Answer 12

To find out whether it's a max or a min, you plug in values for the points around the zero, to see whether it's negative before and positive after (implying a minimum) or a positive before and negative after (implying a maximum).

Answer 13

that's what this question wants you to do - given the interval

0≥x>2

on the function
\[f(x)=\frac{x^4}{4}-x^3+x^2\]
is there a local maximum, local minimum, or both.

Answer 14

Remember, the slope of a line is given by
\[m=\frac{\Delta y}{\Delta x}\]
saying
\[\frac{\mathrm{d}y}{\mathrm{d}x} \]
is exactly like that, only exact.

Answer 15

you could also get this answer without calc. Just look at the roots, and the behavior around the roots. Then you will know if it has a max, min...

Answer 16

Example, x^2-1 has roots at -1, and 1, we know the parabola opens up, so it must have a min between -1, and 1 , because it is continuous.

Answer 17

Last thing before I stop barraging you.

When you said that the max was 0, I think that you meant that the maximum value for that function was 0. Which it is.

I, however, thought that you were talking about the derivative being equal to zero - which happens to be at x=0 for that function. Sorry for that confusion.

If, say, the maximum value for the function *wasn't 0, ie
\[f(x)=-x^2+2\]
the maximum value for the function would be 2. However, the value of the derivative at x=0 would be 0
\[f'(0)=-2*0=0\]

We know the equation of the line because we know both the slope (df/dx=0) and 2 points on the original function (0,2), so can use point slope form

\[(y-y_1)=m(x-x_1)\]
\[(y-2)=0(x-0)\]
\[y=2\]

K, I'm done until you say go ^_^

Answer 18

@zzr0ck3r For this degree 4, since we also know it opens up (since the leading term is positive) how could you tell without calc if it had a max or min on that interval?

Answer 19

@johnny101 I'm sorry I've frightened you away :(

Answer 20

what are the roots?

Answer 21

it has one real root, (0,0)
do you think it has a min?

Answer 22

yes that is a parabola...:)

Answer 23

\[\frac{1}{4}x^2(x^2-4x+4)=0\]
so
\[x=0\]
\[x=2\]

Answer 24

^_^

Answer 25

sorry forgot about the 1/4

Answer 26

ok it has two roots, are those bounces or does it cross over?

Answer 27

which is definitely interval - if they crossed over there would be more than 2 roots methinks

Answer 28

so it bounces