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# finalpracticesolutions - Math 2250 Final Exam Practice...

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Math 2250 Final Exam Practice Problem Solutions 1. What are the domain and range of the function f ( x ) = ln x x ? Answer: x is only deﬁned for x 0, and ln x is only deﬁned for x > 0. Hence, the domain of the function is x > 0. Notice that lim x 0 + ln x x = -∞ , since x 0 + as x 0 + . Now, we can evaluate lim x →∞ ln x x using L’Hˆopital’s Rule; it is equal to lim x →∞ 1 x 1 2 x = lim x →∞ 2 x x = lim x →∞ 2 x = 0 . Therefore, f will have some maximum value; to ﬁgure out what it is, take f 0 ( x ) = x 1 x - ln x 1 2 x x = 2 2 x - ln x 2 x x = 2 - ln x 2 x 3 / 2 . Then f 0 ( x ) = 0 when 0 = 2 - ln x, meaning that ln x = 2, or x = e 2 . Notice that f 0 ( x ) changes sign from positive to negative at x = e 2 , so the maximum of f occurs here. Since f ( e 2 ) = ln e 2 e 2 = 2 e , we see that the range of f is ± -∞ , 2 e ² . 2. Find the inverse of the function f ( x ) = 1000(1 + 0 . 07) x . Answer: To ﬁnd the inverse, switch the roles of x and y , then solve for y: x = 1000(1 . 07) y ; taking the natural log of both sides, we see that ln x = ln(1000(1 . 07) y ) = ln 1000 + ln(1 . 07 y ) = ln 1000 + y ln 1 . 07 . Therefore, y ln 1 . 07 = ln x - ln 1000 . Hence, y = ln x - ln 1000 ln 1 . 07 . 1

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y = e 3 x at which the tangent line passes through the origin. Answer: Let f ( x ) = e 3 x . Since f 0 ( x ) = 3 e 3 x , the tangent line to e 3 x at the point x = a has slope 3 e 3 a ; hence, using the point-slope formula, it is given by y - e 3 a = 3 e 3 a ( x - a ) = 3 e 3 a x - 3 ae 3 a . In other words, the tangent line to the curve at x = a is y = 3 e 3 a x - 3 ae 3 a + e 3 a or y = e 3 a (3 x - 3 a + 1) . This passes through the origin if we get equality when we substitute 0 for both x and y , so it must be the case that 0 = e 3 a (0 - 3 a + 1) = e 3 a (1 - 3 a ) . Since e 3 a 6 = 0, this means that 1 - 3 a = 0, or a = 1 / 3. Therefore, since f (1 / 3) = e 3 · 1 / 3 = e, the point whose tangent line passes through the origin is ± 1 3 ,e ² . 4. Find the equation of the tangent line to the curve xy 3 - x 2 y = 6 at the point (3 , 2). Answer: Diﬀerentiating both sides with respect to x yields y 3 + 3 xy 2 dy dx - 2 xy - x 2 dy dx = 0 . Therefore, dy dx ( 3 xy 2 - x 2 ) = 2 xy - y 3 . Thus, dy dx = 2 xy - y 3 3 xy 2 - x 2 . Plugging in (3 , 2), we see that the slope of the tangent line is 2(3)(2) - 2 3 3(3)(2) 2 - 3 2 = 12 - 8 36 - 9 = 4 27 . Thus, using the point-slope formula, the equation of the tangent line is y - 2 = 4 27 ( x - 3) = 4 27 x - 12 27 , or, equivalently, y = 4 27 x + 14 9 . 2
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finalpracticesolutions - Math 2250 Final Exam Practice...

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