f(x) = -4^3 - 15x^2 + 18x - 5
Write equation of the tangent line at x = -1

- anonymous

f(x) = -4^3 - 15x^2 + 18x - 5
Write equation of the tangent line at x = -1

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- anonymous

@jim_thompson5910 Can you help? c:

- jim_thompson5910

First derive with respect to x
f(x) = -4x^3 - 15x^2 + 18x - 5
f ' (x) = -3*4x^2 - 2*15x^1 + 18
f ' (x) = -12x^2 - 30x + 18

- jim_thompson5910

plug in x = -1 into f ' (x) = -12x^2 - 30x + 18 to find the slope of the tangent line at x = -1
this will give you the value of m

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## More answers

- jim_thompson5910

tell me what you get for m

- anonymous

36

- jim_thompson5910

the slope of the tangent line at x = -1 is m = 36, good

- jim_thompson5910

now tell me what you get when you plug x = -1 into the original function f(x)

- anonymous

-34

- jim_thompson5910

this means the point (-1, -34) lies on the graph of the function f(x)

- jim_thompson5910

you now have a slope and a point...all the necessary ingredients to find the equation of a line

- anonymous

y = 36x + 2

- jim_thompson5910

that's correct

- jim_thompson5910

nice work

- anonymous

Awesome! :O Well, ok. How about this,
Find the x value(s) where the slope of the tangent line is 0

- jim_thompson5910

plug in f ' (x) = 0 and solve for x
f ' (x) = -12x^2 - 30x + 18
0 = -12x^2 - 30x + 18
-12x^2 - 30x + 18 = 0
...
x = ?? or x = ??

- anonymous

OHHHHHHHHHHHHHHHHHHHHHHH -3 and -0.5

- jim_thompson5910

one of those is wrong

- anonymous

\[\frac{ 5\pm \sqrt{5^2 - 4(-2)(3)} }{ 2(-2) }\]

- anonymous

Is there anything wrong with the quadratic formula?

- jim_thompson5910

you should have (5+7)/(-4) and (5-7)/(-4)

- anonymous

Oh I got it, the 0.5 isn't negative

- jim_thompson5910

good

- anonymous

So if they tell me to find the Max/Min points I just use the x values and plug it back in to the original formula to find y, then I'd have the points.

- Jhannybean

mmhmm

- jim_thompson5910

you have to use the first or second derivative test though

- anonymous

So max would be (-0.25,0.5) min would be (-3,-86)

- anonymous

What is that o.o

- Jhannybean

The first derivative tells you where the function is increasing and decreasing using the critical points, or the "zeroes" of a function
The second derivative tells you where the function is concave up or down,and the inflection points where the original function changes concavity.

- anonymous

I'm sorry, but can you dumb it down a bit? Lol

- Jhannybean

lets see...|dw:1370328876443:dw|

- Jhannybean

that's the first part.

- Jhannybean

|dw:1370329332922:dw|

- Jhannybean

hope this helps x.x

- anonymous

LOL I really do appreciate your effort, what's inc and dec and conc, though?

- Jhannybean

increasing, decreasing, concave up/down

- anonymous

Lol I'm sorry

- anonymous

Hey how do I find the point of influence/inflection on this formula?

- anonymous

The inflection points are the points where the second derivative changes sign. That means you want to find the points where the second derivative is equal to 0.

- anonymous

So -24x - 30 = 0?

- anonymous

Yeah, you got it. So what is the inflection point knowing that?

- anonymous

So the point is (-1.25, -43.125) ? ._.

- anonymous

I didn't check if your plugging in the x value is correct, but your x is, so I'll assume you did it properly. Good job! Although I'd feel a lot more comfortable with someone else double checking me ^^"

- anonymous

But doesn't it seem a little weird?

- anonymous

The numbers are all over the place. xD

- anonymous

They can be. A critical point isn't the same as an inflection point. A critical point is when the first derivative changes sign, and inflection is for second derivative. Derivatives simply define how the graph of its integral is changing with respect to a variable. In this case, it shows how y is changing as x changes. And the tangent line is just the extended estimation of any point using the derivative to base it. If any of that doesn't make sense, I'd be happy to clarify further.

- anonymous

No I get it, it's fine. I gotta go to bed anyways. xD It's 1 in the morning

- anonymous

4 am here >.> Good luck on your test! Don't panic, and use common sense before attacking with numbers.

- anonymous

I'll try ;-; Goodnight. You should go to bed early too xD

- anonymous

Ha... "early" has flown out the window already. And I have work tomorrow xD Goodnight!

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