F(x) = x^5 - x^3 + x
f'(x) = 5x^4 - 3x^2 + 1
f(3) = (3)^5 - (3)^3 + 3 = 243 - 27 + 3 = 219
f'(3) = 5(3)^4 - 3(3)^2 + 1 = 5(81) - 3(9) + 1 = 405 - 27 + 1 = 379
Let the required equation of the tangent line be
y = mx + c; where y = 219, m = 379, x = 3
219 = 379(3) + c = 1,137
c = 219 - 1,137 = -918
Therefore, required equation is
y = 379x - 918
Plug in -3 in the x.
4(2)(-3)= -24.
The segment goes from point (-1, 3) to (1,-1)
Slope = rise / run = (-3 - 1) / (1 - (-1)) = -4 / 2 = -2
Slope = -2
ohhhh, what? !?? balance ?
Answer:
Could you repost your question with a picture of the arc?
