Insight Builder Coursework Materials

Insight Builders #2b ANSWERS Due: September 19, 2014 1 . Let f be the function given by f (x) = x 2 − 3x + 2 . (a) Find the domain of f. (b) Find the slope of the secant line joining ( 3, f (3)) and ( 4, f (4)) . (c) Let g(x) = f (x) + 2 . Find the slope of the secant line joining ( 3, g(3)) and ( 4, g(4)) . (d) Let h(x) = f (2x) . Find the slope of the secant line joining (1.5, h(1.5)) and ( 2, h(2)) . (a) x 2 − 3x + 2 ≥ 0 (x − 2)(x − 1) ≥ 0 3 x ≤ 1 or x ≥ 2 f (3) = 9 − 9 + 2 = 2 (b) f (4) = 16 − 12 + 2 = 6 m= Δy = Δx 2 6− 2 = 6 − 2 ≈ 1.035 4−3 (c) slope will be the same as in (b) or g(3) = 2 + 2 g(4) = 6 + 2 m= (d) Δy = Δx ( m= ) ( 6+2 − 4−3 2+2 )= 2 6 − 2 ≈ 1.035 h(2) − h(1.5) f (4) − f (3) = = 2 − 1.5 .5 6− 2 =2 .5 ( 6− 2 ) 2 2. Let h be a function defined for all x ≠ 0 such that h(4) = −3 and the derivative of h is given by x2 − 2 for all x ≠ 0 . h′(x) = x (a) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values. Justify your answers. (b) On what intervals, if any, is the graph of h concave up? Justify your answer. (c) Write an equation for the line tangent to the graph of h at x = 4 . (d) Does the line tangent to the graph of h at x = 4 lie above or below the graph of h for x > 4 . Why? (a) If there is a horizontal tangent then the slope must be zero. - + - + x2 − 2 =0 x 2 x −2=0 x=± 2 h′(x) = 0 The graph of h has a horizontal tangent at x = ± 2 . h has a local maximum at x = 0 , where h′ changes from positive to negative, although the derivative is undefined at x = 0 . h has local minima at x = ± 2 , where h′ changes from negative to positive. x2 − 2 2 = x− x x 2 h′′ = 1 + 2 x b) h′(x) = ( ) h′′ ( 2 ) = 2 h′′ − 2 = 2 + + 0 h′′ is undefined at x = 0 , an positive elsewhere, so h is concave up everywhere except at x = 0 . c) h′(4) = 16 − 2 7 = and (4,−3) is a point on h. 4 2 7 y + 3 = (x − 4) 2 d) Since h is concave up, the tangent line lies below the curve. 3. Solve the following inequalities. Draw number lines as part of your work. 2 a) x > 20 2 b) x − 5x + 2 ≤ −4 x2 − 5 > 2 c) and x 2 − 16 ≤ 0 Solve  the  related  equation:         x 2 = 20 x = ± 20 = ±2 5   Make  a  number  line:           Exclude  the  2  values  of  x  above.       Now  test  a  point  in  each  of  the  3  intervals.    Do  so  by  selected  a  point,  such  as  -­‐10,  then  checking  to  see  if  the  original   inequality  is  true.    If  so,  then  graph  that  interval.  Since   (−10) = 100 > 20 ,  the  region  to  the  left  of   −2 included.    Similarly,  the  center  interval  is  excluded  and  the  interval  to  the  right  is  included.     2 The  final  answer  is   (−∞,− 20 ) ∪ ( 20, ∞)  .     b)     x       2 − 5x + 2 ≤ −4     x 2 − 5x + 6 ≤ 0     ( x − 3)( x − 2) ≤ 0 ( x − 3)( x − 2) = 0 x = 2,3 2 3   [2, 3]         c)     x − 5 > 2 and For  the  first  inequality:     2 x − 16 ≤ 0     2 x 2 − 5 > 4     x2 − 9 > 0 x2 − 9 = 0 x = ±3 -3 3   Now,  for  the  second  inequality:       x 2 − 16 = 0     x = ±4 -4 4   Finally,  putting  both  together:     [− 4, − 3) ∪ (3, 4]  .   -4 -3 34 5    is