Calculus derivitives?
find the tangent line(s) to the curve f(x) = x^(3) - 9x through the point (1,-9) .
.... idk what to do :/
find the tangent line(s) to the curve f(x) = x^(3) - 9x through the point (1,-9) .
.... idk what to do :/
?
Favorite Answer
3x^2 -9
f'(1) = -6
(y+9)=(-6)(x-1)
Ironboxy
f(x) = x(x^2 - 9)
= x(x - 3)(x + 3)
From this you know that the graph will come up from the left and hit (-3, 0) up and over(mountain) to come back down and pass through the origin (0, 0), then curve(valley) back up to hit (3, 0) and up up and away to the far top right.
I did my small drawing of the graph and the point( 1, -9) and to me, the only possible line that could possibly exist as a line and pass through (1,-9) and a point on the graph is the origin...
f(0) = 0
f'(0) = 3(0^2) - 9 = -9
We have (0, 0)<--a point on the curve and the line; (1, -9), a point of the line.
the m = (∆y)/(∆x) of the line is (-9 - 0) / (1 - 0) = -9
And sine the derivative (slope) at the point (0,0) on the graph is -9, we now have a tangent line to the curve f(x) that goes through the point (1, -9)
y - y1 = m(x - x1)
y + 9 = -9(x - 1)
y = -9x + 9 - 9
y = -9x
The line is y = -9x
Hope this helps :D
http://www.wolframalpha.com/input/?i=y+%3D+x%5E3+-+9x
Try drawing your own sketch and logically think though where the possible tangent line might be that could pass though a very far-down point like (1, -9)
brballin
use it's derivative and plug in the x value of the point, which is 1 to get m.
f'(x) = 3x^2-9 =>plug in 1 =>-6 = m
equation for tangent line is
y - y1 = m(x-x1) x1 and y1 are the values of the given point
so final answer is y = -6(x-1)-9
but...f(1) is -8..so idk
?
you could desire to coach the chain rule. This function has 5 nested applications: cos, sqrt, sin, tan, and a linear function (?x). bear in recommendations that cos' = -sin, sqrt' = a million / (2*sqrt), sin' = cos, tan' = sec^2 or a million + tan^2, and (ax)' = a y = cos?(sin(tan?x)) y' = -sin?(sin(tan?x)) * (?(sin(tan?x)))' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * (sin(tan?x))' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (tan?x)' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (a million + tan^2(?x)) * (?x)' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (a million + tan^2(?x)) * ? I leave any needed simplication to you, yet I do see a ?(sin(tan?x)) in the two the numerator and denominator.