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The solutions to quiz 6 of the math 161 course at millersville university. The quiz covers finding the equation of the tangent line to a curve and applying the mean value theorem. The steps for finding the slope of the tangent line and the equation of the tangent line using implicit differentiation and the point-slope formula.
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Millersville University Name Department of Mathematics MATH 161, Quiz 6 February 27, 2004
Please answer the following questions. Your answers will be evaluated on their correctness, completeness, and use of mathematical concepts we have covered. Please show all work and write out your work neatly. Answers without supporting work will receive no credit.
d dx
[ x^3 y^3
d dx
[9y]
3 x^2 y^3 + x^3
d dx
[ y^3
] = 9 y′(x)
3 x^2 y^3 + x^33 y^2 y′(x) = 9 y′(x) 3 x^2 y^3 = − 3 x^3 y^2 y′(x) + 9y′(x) 3 x^2 y^3 =
( − 3 x^3 y^2 + 9
) y′(x)
x^2 y^3 =
( −x^3 y^2 + 3
) y′(x) x^2 y^3 3 − x^3 y^2
= y′(x)
Thus the slope of the tangent line at the point with coordinates (x, y) = (1, 3) is
m = y′(1) =
Using the point-slope formula for a line we get as the equation of the tangent line:
m =
y − y 1 x − x 1
−
y − 3 x − 1 −
(x − 1) = y − 3
−
x +
= y − 3
−
x +
−
x +
= y
f ′(c) =
f (1) − f (0) 1 − 0
Since f ′(x) = 3x^2 + 2x we need to solve the equation
3 c^2 + 2c = 2 3 c^2 + 2c − 2 = 0
c =
√ 22 − 4(3)(−2) 2(3)
=
Since −^1 −
√ 7 3 <^ 0 we will ignore this solution.^ The^ c^ promised by the Mean Value Theorem is c =
Note that 0 < c < 1.