Consider the function f(x) = 5x2 + 3. Determine the point at which the tangent to the curve f(x) at the point x = -2 intersects the function g(x) = x.
I got no idea for this one :|
At x = -2 f(x) = 5 (-2^2) + 3 = 23 y = 23
The slope of f(x) is the derivative 10x and at -2 this equals -20
So we have a point -2,23 and slope = -20
y=mx+b form of this line 23 = -20(-2) + b or b = -17
so the tangent line is y = -20x-17 Now just set this equal to g(x)
g(x) = x x = -20x-17 21x=-17 x = -17/21 = -.81
So the point of intersection is -.81 , -..81
Here is a graph:
Isn't the derivative of f(x)=5x^2+3 = f'(x)=10x
So the Point of Intersection would be: (-17/21,-17/21)
Since the actual slope is -20
Setting them equal to one another
x=-17/21 -- Plugging in to find y
POI = (-17/21,-17/21)Is that correct?
You are correct Julius ...I got the derivative of 5x^2 + 3x.....
sorry the solution methodology is the same though ...I will correct it.
Yep! No need to re-do it, I copied your steps with the correct derivative. Thanks :)
...and you got the correct answer! Good job, Julius. Thanx for finding my error...I corrected it anyway.