find the point of (x^2+4)y=8 which has a horizontal tangent line (use implicit to find dy/dx)?
回答 (3)
2018-11-26 12:38 pm
Differentiating each side wrt x [recalling y = y(x)] gives :
2x * y + (x² + 4)*y' = 0.
A horizontal tangent is when y'=0. Put this condition in the above eqs. Conclusions?
2x * y + (x² + 4)*y' = 0.
A horizontal tangent is when y'=0. Put this condition in the above eqs. Conclusions?
2018-11-26 1:13 pm
(x²+4)y = 8
Differentiate implicitly:
(2x+0)y + (x²+4) dy/dx = 0
(x²+4) dy/dx = −2xy
dy/dx = −2xy/(x²+4)
Curve has horizontal tangent when dy/dx = 0
−2xy/(x²+4) = 0
x = 0 or y = 0
When y = 0 we get:
(x²+4)0 = 8
0 = 8
Not possible
When x = 0 we get:
(0²+4)y = 8
y = 2
Horizontal tangent at point (0,2)
2018-11-26 12:55 pm
We have y = 8/(x^2+4)
dy/dx = -16x/(x^2+4)^2
dy/dx = 0 if x = 0; hence, y = 8/4=2
(0,2) is such point.
dy/dx = -16x/(x^2+4)^2
dy/dx = 0 if x = 0; hence, y = 8/4=2
(0,2) is such point.
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