Transcription of Answers/Solutions to Assigned Even Problems
1 Answers/Solutions to Assigned even : Find the difference quotient forf(x) = 2x+ :Recall that thedifference quotientof a function is:f(x+h) f(x)hForf(x) = 2x+ 3 we have:f(x+h) f(x)h=[2(x+h) + 3] [2x+ 3]h=2x+ 2h+ 3 2x 3h= 2x+ 2h+ 3 2x 3h=2hh= : Find the difference quotient forf(x) = :2x+ :Answer:x-intercept = -4,y-intercept = , slope = :Answer:x-intercept = ,y-intercept = -3, slope = : Find the equation of the line through ( 1,2) with :y=23x+ : Find the equation of the line through (0,0) with slope :y= : Find the equation of the line through (2,5) which is parallel to :x= 2 (parallel to theyaxis means vertical) : Find the equation of the line through (2,5) and (1, 2).
2 Solution:First find the slope:m=( 2) (5)(1) (2)= 7 1= 7 Method 1:Using point-slope form:Now that we have the slope, we can use point-slope form with (x0, y0) = (2,5)andm= 7:y y0=m(x x0)y (5) = (7)(x (2))y 5 = 7x 14y= 7x 9 Method 2:Using slope-intercept form:Since we know the slope,m= 7, slope-intercept form gives:y=mx+b= y= 7x+bWe can then findbby plugging in either point into this equation. For example,plugging in (1, 2), we get:( 2) = 7(1) +b= b= 9 Hence, we get:y= 7x : Find the equation of the line through ( 2,3) and (0,5).
3 Answer:y=x+ : Find the equation of the line through (1,5) and (1, 4).Answer:x= : : Find the slope of the tangent line tof(x) = 3 atx= :Recall that the slope of the tangent line atx= 1 is defined by:slope of tangent atx= 1=limh 0f(1 +h) f(1)hNow,f(x) = 3 is aconstantfunction, ,f(anything) = 3. So we have: f(1) = 3 f(1 +h) = 3So the slope formula gives:slope of tangent line atx= 1= limh 0f(1 +h) f(1)h= limh 0[ 3] [ 3]h= limh 0(0h)= limh 0(0)= 0 See the next page for an explanation this s the picture for what s going on in problem :The functionf(x) = 3 is a constant function.
4 Hence, its graph is a horizontal line,which has slopem= (x) = 3 The tangent line to this graph atx= 1 is the line that touches the graph at (1, f(1))and lays flattest against the graph there. Hence, in this case, the tangent line is theoriginal line itself, and thus also has slopem= atx= : Find the slope of the tangent line tof(x) = 2 7xatx= : Find the slope of the tangent line tof(x) =x2 1 atx= : Letf(x) = 2x ) Compute the slope of the secant line joining the points wherex= 0 andx= :Recall that, in general, the secant line to a functionf(x) throughx=aandx=bis the (unique) line that goes through (a, f(a)) and (b, f(b)):slope of secant =f(b) f(a)b a(a, f(a))(b, f(b))y=f(x)secantabSolution.
5 Hence, forf(x) = 2x x2, the secant line throughx= 0 andx=12is:f(12) f(0)12 0=[ 2(12) (12)2] [ 2(0) (0)2]12=[ 1 14] [ 0 ]12=3412=34 21=64=325b) Use calculus to compute the slope of the tangent line tof(x) atx= 0, and comparewith the slope found in part a).Solution:Sincef(x) = 2x x2, we have:slope of tangent line atx= 0= limh 0f(0 +h) f(0)h= limh 0f(h) f(0)h= limh 0[2(h) (h)2] [2(0) (0)2]h= limh 02h h2h= limh 0h(2 h)h= limh 0 h(2 h) h= limh 0(2 h)= 2(As usual, we can think of the last step as simply plugging inh= 0.)Note that this is close to the slope found in part a).
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