Write and equation of the tangent line y=f (x) to the function at the given value.
F (x)= (x^2+7x+4)/(2x), x=-3
I got y= 1/2x + 1/6
Everybody in my class is getting different answers and the teacher is just saying to figure it out, so please help.
Write and equation of the tangent line y=f (x) to the function at the given value.?
2016-05-05 4:09 pm
回答 (2)
2016-05-05 4:30 pm
y = f(x) = (x² + 7x + 4) / (2x)
f '(x) = [(2x)(x² + 7x + 4)' - (x² + 7x + 4)(2x)'] / (2x)²
f '(x) = [(2x)(2x + 7) - (x² + 7x + 4)(2)] / (4x²)
f '(x) = (4x² + 14x - 2x² - 14x - 8) / (4x²)
f '(x) = (2x² - 8) / (4x²)
f '(x) = (x² - 4) / (2x²)
Slope of the tangent at (x = -3)
= f '(-3)
= [(-3)² - 4] / [2(-3)²]
= 5/18
When x = -3 :
y = [(-3)² + 7(-3) + 4] / [2(-3)]
y = 4/3
The slope of the tangent is 5/18, and the point of contact is (-3, 4/3).
The equation of the tangent is :
[y - (4/3)] = (5/18) [x - (-3)]
[y - (4/3)] (18) = (5/18) (x + 3) (18)
18y - 24 = 5(x + 3)
18y - 24 = 5x + 15
5x - 18y + 39 = 0
f '(x) = [(2x)(x² + 7x + 4)' - (x² + 7x + 4)(2x)'] / (2x)²
f '(x) = [(2x)(2x + 7) - (x² + 7x + 4)(2)] / (4x²)
f '(x) = (4x² + 14x - 2x² - 14x - 8) / (4x²)
f '(x) = (2x² - 8) / (4x²)
f '(x) = (x² - 4) / (2x²)
Slope of the tangent at (x = -3)
= f '(-3)
= [(-3)² - 4] / [2(-3)²]
= 5/18
When x = -3 :
y = [(-3)² + 7(-3) + 4] / [2(-3)]
y = 4/3
The slope of the tangent is 5/18, and the point of contact is (-3, 4/3).
The equation of the tangent is :
[y - (4/3)] = (5/18) [x - (-3)]
[y - (4/3)] (18) = (5/18) (x + 3) (18)
18y - 24 = 5(x + 3)
18y - 24 = 5x + 15
5x - 18y + 39 = 0
2016-05-05 4:12 pm
"I see it". y= 1-:-2x + 1-:-6 "So do know".
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