Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Calculus 1 Worksheet #13 Even Review for Test #3 2 f ( x ) = 3 x − 4 ⇒⇒ f '( x ) = lim h→0
4
f(x)=
3( x + h) − 4 − (3 x − 4) 3 x + 3h − 4 − 3 x + 4 3h = lim = lim = lim 3 = 3 h → 0 h → 0 h h h h→0
2 x+4
2( x + 4) − 2 ( ( x + h ) + 4 ) ( x + h) + 4 2 2 2 ( x + 4) 2 − • − • ( x + h ) + 4 x + 4 = lim ( x + h ) + 4 ( x + 4) x + 4 ( x + h ) + 4 = lim = ( x + h + 4 ) ( x + 4) = f '( x) = lim h →0 h →0 h →0 h h h 2 x + 8 − 2 x − 2h − 8 −2 h −2 −2 = lim = lim = lim = h → 0 h ( x + h + 4 ) ( x + 4) h → 0 h ( x + h + 4 ) ( x + 4) h → 0 ( x + h + 4 ) ( x + 4) ( x + 4) 2
6 8 10
1 −1 d d 1 3 − 2 x = ( 3 − 2 x ) 2 = ( 3 − 2 x ) 2 • ( −2 ) = dx dx 2
2
1
y = 3x 3 − 4 x 2 − 2 →
−1 3 − 2x
−1 −1 −1 −1 dy 2 1 = • 3x 3 − • 4 x 2 = 2 x 3 − 2 x 2 dx 3 2
−1 ⎡ −2 x ⎤ ⎡1 ⎤ 1 − x2 x2 1 − x 2 (1) − x ⎢ 1 − x 2 (1) − x ⎢ (1 − x 2 ) 2 ( −2 x ) ⎥ + ⎥ 2 1 x dy 1 − x2 = ⎣2 ⎦= ⎣ 2 1− x ⎦ = ⇒⇒ = y= 2 2 1 − x2 dx 1 − x2 1 − x2 1 − x2
)
(
(
)
1 − x2 + x2 1 − x2 1 − x2
12
14 16
=
1
(1 − x ) 2
1 − x2
or
1
(1 − x )
2 3
1 ⇒⇒ y ' = − x −2 →→ y '' = 2 x −3 →→ y ''' = −6 x −4 →→ y iv = 24 x −5 →→ y iv (1) = 24(1) −5 = 24 x y = 5→ y' = 0 y=
x x x 2x 3x ⎛ 1 ⎞ f ( x) = x x ⇒⇒ f '( x) = x ⎜ + →→ + →→ → ⎟ + x (1) →→ 1 2 x 2 x 2 x 2 x ⎝2 x ⎠ 3 3x 3x x 3 x 3 1 3 x • → = or smarter! f ( x) = x x = x 2 → f '( x) = x 2 = x 2x 2 2 2 2 x x
18 20 22
y = ( x 2 − 2 )( x −1 + 2 ) ⇒⇒
d 2 x − 2 )( x −1 + 2 ) = ( x −1 + 2 ) ( 2 x ) + ( x 2 − 2 )( − x −2 ) = 2 + 4 x − 1 + 2 x −2 = 1 + 2 x −2 + 4 x ( dx 2 5 d 2 5 2 10 y = 3x 2 + − 2 ⇒⇒ (3 x 2 + − 2 ) = 6 x − 2 + 3 x x dx x x x x
3( x − 1) 2 ⎛ 2 ⎞ 6( x − 1) 2 ⎛ x −1 ⎞ ⎛ x − 1 ⎞ ⎛ ( x + 1)(1) − ( x − 1)(1) ⎞ f ( x) = ⎜ ⎟ → f '( x) = ⎜ ⎟ → f '( x) = ⎟ ⇒⇒ f '( x) = 3 ⎜ ⎟ ⎜ ( x + 1) 2 ( x + 1) 2 ⎝ ( x + 1) 2 ⎠ ( x + 1) 4 ⎝ x +1⎠ ⎝ x +1⎠ ⎝ ⎠ 24 Find the x and y intercepts of the line that is tangent to the curve y = x3 at the point (-2,-8) 3
2
Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Calculus 1 Worksheet #13 Even Review for Test #3 x − int@(−4 / 3, 0) y = x 3 ⇒⇒ y ' = 3 x 2 → y '(−2) = 12 → y + 8 = 12( x + 2) → y = 12 x + 16 → y − int@(0,16)
26
1 Find the slope of the normal to f(x) =2x3+x2-1 at the point x= . 2 5 −2 ⎛1⎞ ⎛1⎞ ⎛1⎞ 3 f '( x) = 6 x 2 + 2 x ⇒⇒ f ' ⎜ ⎟ = 6 ⎜ ⎟ + 2 ⎜ ⎟ = + 1 = Slope of the normal = 2 5 ⎝2⎠ ⎝4⎠ ⎝2⎠ 2 28 1 then find and graph the tangent at point (3, f(3)) Graph the function whose equation is g(x)= x−2 1 −1 ⇒⇒ g '( x) = → g '(3) = −1 → y − 1 = −1( x − 3) → y = − x + 4 g ( x) = x−2 ( x − 2) 2 30 Graph the function y = 2 x 2 + 4 x − 1 then find and graph the tangent line at the point where x=1. y = 2 x 2 + 4 x − 1 ; tangent at x=1
y = 2( x 2 + 2 x + 1) − 1 − 2 → y = 2( x + 1) 2 − 3 → y ' = 4 x + 4 → y '(1) = 8 → y − 5 = 8( x − 1) → y = 8 x − 3 32 Find the points on the curve x + 1 where the tangent is parallel to the y-axis. 1 y = x + 1 ⇒⇒ y ' = ⇒⇒ undefined @(−1, 0) 2 x +1 34 dy for y = ( x − 1) ( x 2 + 5 x − 1) then find the slope of the tangent when x=3 Find dx dy dy y = ( x − 1)( x 2 + 5 x − 1) ⇒⇒ = ( x − 1)(2 x − 5) + ( x 2 + 5 x − 1)(1) ⇒ = 2 x2 + 5x − 2 x − 5 + x2 + 5x − 1 dx dx dy dy = 3x 2 + 8 x − 6 ⇒⇒ (3) = 27 + 24 − 6 = 45 dx dx 36 Find the slope of the tangent line and the normal line to the curve y = x x + 1 at the point where x=1.
dy dy x dy 1 2 4 2 5 2 ⎡ 1 ⎤ = x⎢ + x + 1(1) → = + x + 1 → (1) = + 2 →→ + = ⎥ dx dx 2 x + 1 dx 4 4 4 2 2 ⎣ 2 x +1 ⎦ 5 2 −2 2 The slope of the normal = 4 5 38 If the slope of a tangent line is 5 then what is the slope of the normal line to the same curve at the same point? −1 5 −2 40 ⎛ x −1 ⎞ Find the equation of the normal line to the curve f ( x) = ⎜ ⎟ at the point where x=2. ⎝ x +1⎠
The slope of the tangent =
−2
⎛ x −1 ⎞ ⎛ x +1⎞ ⎛ x + 1 ⎞ ⎛ ( x − 1)(1) − ( x + 1)(1) ⎞ ⎛ x + 1 ⎞ ⎛ −2 ⎞ f ( x) = ⎜ ⎟ = 2⎜ ⎟ ⇒⇒ f ( x) = ⎜ ⎟ ⇒ f '( x) = 2 ⎜ ⎟⎜ ⎟⎜ 2 2 ⎟ ( x − 1) ⎝ x +1⎠ ⎝ x −1 ⎠ ⎝ x −1 ⎠ ⎝ ⎝ x − 1 ⎠ ⎝ ( x − 1) ⎠ ⎠ =
2
1 1 −4( x + 1) −4(3) ⇒ f '(−2) = = −12 ⇒ Slope of normal = ⇒ y − 9 = ( x − 2) 3 ( x − 1) 1 12 12