Version PREVIEW HW 04 hoffman (57225) 1. Consequently,

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1 Version PREVIEW HW 04 hoffman (57225) 1 This print-out shoul have 12 questions Multiple-choice questions may continue on the next column or page fin all choices before answering CalCa02b points Let f be the function efine by lim h 0 f(5 + h) f(5) h = 27 Determine if f(x) = x + (x 2 + x 2 ) 2 lim h 0 f(5 + h) f(5) h exists, an if it oes, fin its value 1 limit = 2 keywors: limit, absolute value, Newtonian quotient, CalCb0a points If f is a function on ( 8, 8) having 2 limit oesn t exist limit = 24 4 limit = 2 5 limit = 25 limit = 27 Since we see that v = x 2 + x 2 = Thus f(x) = { { v, v 0, v, v < 0, { 2(x 2), x 2, 0, x < 2 x, x < 2, x + 4(x 2) 2, x 2 ; in particular, f(5) = 41, while as its graph, which of the following is the graph of the erivative of f? 1 f(5 + h) = 5 + h + 4( + h) 2 = h + h 2 for small values of h (so that 5 + h > 2) in this case, f(5 + h) f(5) = h(27 + h)

2 Version PREVIEW HW 04 hoffman (57225) On ( 8, 8) f is a continuous function which is ifferentiable everywhere except at x = 2 an x =, so f (x) is not efine at these points This eliminates immeiately two of the graphs On the other han, f has critical points at x = an x = 0 as well as at x = 2, x = Thus f (x) = 0 at x =, 0 But both of the remaining graphs have x- intercepts, so this oesn t help us However, f is ecreasing on ( 8, ), so the graph of f must lie below the x-axis on ( 8, ); similarly, f is increasing on (, 0), so the graph of f must lie above the x-axis on (, 0) the graph of f must be 4 CalCc02s points Determine the erivative of f when f(x) = ( ) 2/ f (x) = 4 15 x 1/

3 Version PREVIEW HW 04 hoffman (57225) 2 f (x) = ( ) 2 x 1/ 5 which after simplification becomes y = 8x + 1 f (x) = 0 4 f (x) oes not exist 5 f (x) = 2 ( ) 1/ 2 5 The erivative of any constant function is zero f (x) = 0 CalCc0a points Fin the y-intercept of the tangent line at the point P ( 1, f( 1)) on the graph of 1 y-intercept = 8 2 y-intercept = 7 f(x) = x 2 2x + 4 y-intercept = 7 4 y-intercept = 8 5 y-intercept = 1 y-intercept = 1 The slope, m, of the tangent line at the point P ( 1, f( 1)) on the graph of f is the value of the erivative f (x) = x 2 the tangent line at P has y-intercept = 1 CalCc10a points Fin the erivative of f(x) = x 1 4 x f (x) = x1 2 + x f (x) = x x 4 f (x) = x1 2 4x 4 4 f (x) = x1 2 4x f (x) = x x 5 4 Since x (xr ) = rx r 1, we see that f (x) = 1 4 ( 1 x 4 + ) x 5 4 at x = 1, ie, m = 8 On the other han, f( 1) = 9 Thus by the point-slope formula, an equation for the tangent line at P ( 1, f( 1)) is y 9 = 8(x + 1), f (x) = x x 5 4 CalCc27n points

4 Version PREVIEW HW 04 hoffman (57225) 4 Fin the value of f (2) when f(x) = x2 + 1 x 1 Correct answer: 0501 Using the Quotient Rule an the fact that we obtain (x α ) = αx α 1, f (x) = 2x(x 1) x 2 (x 2 + 1) (x 1) 2 At x = 2, therefore, Fin p(x) so that x f (2) = 2 49 CalCc29a points { x 2 2x + 2 } 2x 2 x + 4 = 1 p(x) = (x 2 + 2) 2 p(x) = (x 2 + 2) p(x) = (x 2 2) 4 p(x) = (x 2 + 4) 5 p(x) = (x 2 + 4) p(x) = (x 2 2) By the Prouct Rule, x { x 2 2x + 2 } 2x 2 x + 4 = p(x) (2x 2 x + 4) 2 2x 2 2x 2 x + 4 (4x 1)(x2 2x + 2) (2x 2 x + 4) 2 After simplification this becomes x { x 2 2x + 2 } 2x 2 = (x2 2) x + 4 (2x 2 x + 4) 2 p(x) = (x 2 2) CalCc8s points Fin equations for those tangent lines to the graph of y = x + 2 x that are parallel to the line y + 2x = 5 1 y + 2x = 0, y + 2x + = 0 2 y + 2x 5 = 0, y + 2x + = 0 y 2x = 0, y 2x + 1 = 0 4 y 2x 5 = 0, y + 2x + 1 = 0 5 y + 2x = 0, y 2x + = 0 y + 2x 5 = 0, y + 2x + 1 = 0 It s easier to ivie first before ifferentiating For then so that y = x + 2 x y = 2 x 2 = x, On the other han, the line y + 2x = 5 has slope 2 Thus a tangent line will be parallel to y + 2x = 5 when y = 2 x 2 = 2

5 Version PREVIEW HW 04 hoffman (57225) 5 Solving for x we see that x = ±1 Hence the tangent line to the graph of y = x + 2 x is parallel to y + 2x = 5 at the points P = (1, ) an Q = ( 1, 1) on the graph But, by the point slope formula, an equation for the tangent line at P is y = 2(x 1), while an equation for the tangent line at Q is y + 1 = 2(x + 1) after simplification these become y + 2x 5 = 0, y + 2x + = 0 CalCe10b points But cos 2 x + sin 2 x = 1, so When the graph f (x) = 5 cos x 2 (5 2 cosx) 2 CalCc77a points f(x) = 1 x + x 2 x + 2 P Fin the erivative of f when f(x) = sin x 5 2 cosx 1 f (x) = 5 sin x 2 cos2 x (5 2 cos x) 2 2 f (x) = 5 cos x 2 (5 2 cosx) 2 f (x) = 5 cos x + 2 (5 2 cosx) 2 4 f (x) = 5 sinx cosx 5 f (x) = 5 cos x 2 sin2 x (5 2 cos x) 2 f (x) = 5 cos x cosx By the Quotient Rule, f (x) = cos x(5 2 cosx) 2 sin x sin x (5 2 cos x) 2 = 5 cos x 2 (cos2 x + sin 2 x) (5 2 cosx) 2 Q of f shows that there are two points P an Q at which the tangent line is horizontal Fin the slope of the straight line, shown in gray above, passing through P an Q 1 slope = 19 2 slope = 10 slope = 4 slope = 7 5 slope = 8

6 Version PREVIEW HW 04 hoffman (57225) The tangent line to the graph of f when f (x) = 0, ie, when f (x) = x x = (x + )(x 1) = 0 Thus the x-coorinates of P an Q are given respectively by x = an x = 1 Thus the points P, Q are given by P = (, f()), Q = (1, f(1)), in which case the line passing through P, Q has slope = But f(1) f() 1 () = f() = 11, f(1) = 1 f(1) f() 4 the line passing through P, Q has slope = 8 CalCh01a points Determine F (x) when F(x) = xf (x) + 2f(x) an f is a twice-ifferentiable function 1 F (x) = 5 f (x) + x f (x) 2 F (x) = f (x) 2 x f (x) F (x) = f (x) x f (x) 4 F (x) = 5 f (x) 2 x f (x) 5 F (x) = f (x) + 2 x f (x) F (x) = 5 f (x) x f (x) After ifferentiation using the Prouct Rule we see that F (x) = f (x) xf (x) TrigDeriv12a points Fin the erivative of f(x) = x 4 (tanx + sec x) 1 f (x) = x (4 x sec x)(tanx sec x) 2 f (x) = x (x sec x 4)(tanx + sec x) f (x) = x (x sec x 4)(tanx sec x) 4 f (x) = x (4 x sec x)(tanx + sec x) 5 f (x) = x (4 + x sec x)(tanx + sec x) f (x) = x (x sec x + 4)(tanx sec x) By the Prouct Rule, f (x) = 4x (tanx + sec x) But then + x 4 (sec 2 x + sec x tanx) f (x) = 4x (tanx + sec x) + x 4 sec x(sec x + tanx) f (x) = x (4 + x sec x)(tanx + sec x) keywors: F (x) = f (x) xf (x) + 2f (x)