anonymous
  • anonymous
f(x) = -4^3 - 15x^2 + 18x - 5 Write equation of the tangent line at x = -1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@jim_thompson5910 Can you help? c:
jim_thompson5910
  • 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
  • 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
  • jim_thompson5910
tell me what you get for m
anonymous
  • anonymous
36
jim_thompson5910
  • jim_thompson5910
the slope of the tangent line at x = -1 is m = 36, good
jim_thompson5910
  • jim_thompson5910
now tell me what you get when you plug x = -1 into the original function f(x)
anonymous
  • anonymous
-34
jim_thompson5910
  • jim_thompson5910
this means the point (-1, -34) lies on the graph of the function f(x)
jim_thompson5910
  • jim_thompson5910
you now have a slope and a point...all the necessary ingredients to find the equation of a line
anonymous
  • anonymous
y = 36x + 2
jim_thompson5910
  • jim_thompson5910
that's correct
jim_thompson5910
  • jim_thompson5910
nice work
anonymous
  • anonymous
Awesome! :O Well, ok. How about this, Find the x value(s) where the slope of the tangent line is 0
jim_thompson5910
  • 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
  • anonymous
OHHHHHHHHHHHHHHHHHHHHHHH -3 and -0.5
jim_thompson5910
  • jim_thompson5910
one of those is wrong
anonymous
  • anonymous
\[\frac{ 5\pm \sqrt{5^2 - 4(-2)(3)} }{ 2(-2) }\]
anonymous
  • anonymous
Is there anything wrong with the quadratic formula?
jim_thompson5910
  • jim_thompson5910
you should have (5+7)/(-4) and (5-7)/(-4)
anonymous
  • anonymous
Oh I got it, the 0.5 isn't negative
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • 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
  • Jhannybean
mmhmm
jim_thompson5910
  • jim_thompson5910
you have to use the first or second derivative test though
anonymous
  • anonymous
So max would be (-0.25,0.5) min would be (-3,-86)
anonymous
  • anonymous
What is that o.o
Jhannybean
  • 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
  • anonymous
I'm sorry, but can you dumb it down a bit? Lol
Jhannybean
  • Jhannybean
lets see...|dw:1370328876443:dw|
Jhannybean
  • Jhannybean
that's the first part.
Jhannybean
  • Jhannybean
|dw:1370329332922:dw|
Jhannybean
  • Jhannybean
hope this helps x.x
anonymous
  • anonymous
LOL I really do appreciate your effort, what's inc and dec and conc, though?
Jhannybean
  • Jhannybean
increasing, decreasing, concave up/down
anonymous
  • anonymous
Lol I'm sorry
anonymous
  • anonymous
Hey how do I find the point of influence/inflection on this formula?
anonymous
  • 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
  • anonymous
So -24x - 30 = 0?
anonymous
  • anonymous
Yeah, you got it. So what is the inflection point knowing that?
anonymous
  • anonymous
So the point is (-1.25, -43.125) ? ._.
anonymous
  • 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
  • anonymous
But doesn't it seem a little weird?
anonymous
  • anonymous
The numbers are all over the place. xD
anonymous
  • 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
  • anonymous
No I get it, it's fine. I gotta go to bed anyways. xD It's 1 in the morning
anonymous
  • anonymous
4 am here >.> Good luck on your test! Don't panic, and use common sense before attacking with numbers.
anonymous
  • anonymous
I'll try ;-; Goodnight. You should go to bed early too xD
anonymous
  • anonymous
Ha... "early" has flown out the window already. And I have work tomorrow xD Goodnight!

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