Analytic Geometry and Calculus I Exam 1 Practice Problems Solutions /19/7 Question 1 Write the following as an integer: log 4 (9)+log (5) We have: log 4 (9)+log (5) = ( log 4 (9)) ( log (5)) = 5 ( log (9)) 4 ( = 5 4 1 ) log4 (9) = 5 ( 4 log 4 (9)) 1 Here we used the properties: = 5(9) 1 = 5() = 15. a b+c = a b a c, a log a (b) = b, (a b ) c = a bc = (a c ) b.
Question Eliminate the parameter t to find a Cartesian equation of the curve given parametrically by the relations: x = 1 + t, y = t and sketch the curve. Also sketch the part of the curve for which t 0. We have: x = 1 + t, x 1 = t, t = 1 (x 1), y = t ( 1 = (x 1) = 1 9 (x 1), ) 9y = 18 (x 1) = 18 (x x + 1) = 17 + x x. So the curve has the equation x x + 9y = 17. This is the equation of a concave down parabola. The equation y = 1 9 (x 1) shows that the parabola has vertex at (1, ) and is symmetrical around the vertical line x = 1. The point (1, ) gives the global maximum for y. As t varies x ranges once through all possible real values, so the whole parabola is traced. The parabola intersects the y-axis at t = 1 (, so at the point 0, 17 9 The parabola intersects the x-axis at t = ±, so at the points ). ( 1 ±, 0 When t = 0, we are at the vertex (1, ). As t increases from zero, the whole right half of the parabola is traced out. ).
Question Find the following limits: lim x x x 4 When x <, we have x < 0, so x = (x ). So we have: lim x + x x 4 x lim x x 4 (x ) x (x )(x + ) x 1 x + x 1 + = 1 4. When x >, we have x > 0, so x = x. So we have: lim x x x 4 Since lim x + x x 4 lim x lim x + x 4 x x + (x )(x + ) 1 x + x + 1 x + + = 1 4. x x x 4 x, the limit lim x x 4 does not exist.
Question 4 Suppose that: f(1) =, f (1) = 4, f() = 4, f () =, g(1) =, g (1) = 5, g() = 1, g () = 7. Let h(x) = f(g(x)), k(x) = f(x)g(x) and m(x) = f(x) g(x). Find the equations of the tangent and normal lines to the curves y = h(x), y = k(x) and y = m(x) at the points with x = 1. We have h(1) = f(g(1)) = f() = 4 and h (x) = f (g(x))g (x), so h (1) = f (g(1))g (1) = f ()(5) = (5) = 10. So the tangent line to y = h(x) has slope 10 and goes through the point (1, 4), so has the equation: y 4 = 10(x 1) = 10x + 10, y + 10x = 14. Also the normal line to y = h(x) has slope 1 10 point (1, 4), so has the equation: and goes through the y 4 = 1 (x 1), 10y 40 = x 1, x 10y = 9. 10 We have k(1) = f(1)g(1) = () = 6 and k (x) = f (x)g(x)+f(x)g (x), so k (1) = f (1))g(1) + f(1)g (1) = 4() + (5) = 1 + 10 =. So the tangent line to y = k(x) has slope and goes through the point (1, 6), so has the equation: y 6 = (x 1) = x, y x = 16. Also the normal line to y = k(x) has slope 1 point (1, 6), so has the equation: and goes through the y 6 = 1 (x 1), y 1 = (x 1) = x + 1, x + y = 1. 4
We have m(1) = f(1) g(1) = and m (x) = f (x)g(x) f(x)g (x), g (x) so m (1) = f (1)g(1) f(1)g (1) 4() (5) 1 10 = = = g (1) 9 9. So the tangent line to y = m(x) has slope and goes through the point ( 9 1, ), so has the equation: y = (x 1), 9y 6 = (x 1) = x, x 9y = 4. 9 Also the normal line to y = m(x) has slope 9 ( point 1, ), so has the equation: and goes through the y = 9 (x 1), 6y 4 = 7(x 1) = 7x+7, 7x+6y = 40. 5
Question 5 Let f(x) = (x 1) x 1 + x. Use the definition f f(a + h) f(a) (a) to find f (1). h 0 h We have f(1) = (1 1) 1 1 + 1 = 0() ) = 0. Then we get: f (1) h 0 f(1 + h) f(1) h f(1 + h) h 0 h (1 + h 1) 1+h h 0 h 1 + (1 + h) h 0 (h) 1+h h h 0 1+h 1 + (1 + h) f(1 + h) 0 h 0 h 1 + (1 + h) = 1+0 1 + (1 + 0) = 1 1 + 1 = 1 + 1 =. 6
Question 6. Let f(x) = xe x. Find f (x) and f (x). f (x) = 1e x + xe x ( x ) = e x x e x = (1 x )e x, f (x) = xe x + (1 x )e x ( x ) = xe x x(1 x )e x = e x ( x x + x ) = e x (x x) = x(x )e x. Find the intervals where f is increasing and the intervals where f is decreasing. When x 1, we have f 0, so f is decreasing on [1, ). When x 1, we have f 0, so f is decreasing on (, 1]. When 1 x 1, we have f 0, so f is increasing on [ 1, 1]. Find intervals where f is concave up and concave down respectively. When x, we have f 0, so f is concave up on [, ). When 0 x, we have f 0, so f is concave down on [0, ]. When x 0, we have f 0, so f is concave up on [, 0]. When x, we have f 0, so f is concave down on (, ]. 7
Sketch the curve y = f(x). As x, the curve approaches the x-axis from below. As x increases, the curve decreases and is initially concave down, switching to concave up at the inflection point (, e ). It then decreases further to its absolute minimum at ( 1, e 1 ). Then it increases, crossing the axes at the origin, where it inflects to concave down. It increases further to its absolute maximum at the point (1, e 1 ). Then it decreases, switching to concave up at the inflection point (, e ). Then it decreases further, approaching the y-axis from above as x. Give the domain and range of the function f. The domain is all the real numbers. The range is from its absolute minimum to its absolute maximum, so is the closed interval [ e 1, e 1 ]. Is f one-to-one? Explain your answer. f is not one-to-one because it has an absolute maximum and in the neighorhood of the absolute maximum it decreases on both sides. Alternatively, we may say that y takes all values in the interval [0, e 1 ] as x ranges over the interval [0, 1], by the intermediate value theorem and all values in the interval (0, e 1 ] over the interval [1, ), again by the intermediate value theorem, so takes all values in the interval (0, e 1 ) exactly twice, once in the interval (0, 1) and once in the interval (1, ). In particular f is not one-to-one. 8