{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW3 - Section 3.1 1 f(x = 0(Rule 1 3 f(x = 5x51 = 5x4(Rule...

This preview shows pages 1–5. Sign up to view the full content.

Section 3.1 1. f ( x ) = 0. (Rule 1) 3. f ( x ) = 5 x 5 - 1 = 5 x 4 . (Rule 2) 5. f ( x ) = 2 . 1 x 2 . 1 - 1 = 2 . 1 x 1 . 1 . (Rule 2) 7. f ( x ) = (3)( x 2 ) = (3)(2 x 2 - 1 ) = 6 x 1 = 6 x. (Rules 2 and 3) 9. f ( r ) = ( π )( r 2 ) = ( π )(2 r 2 - 1 ) = 2 πr 1 = 2 πr. (Rules 2 and 3) 11. f ( x ) = (9)( x 1 / 3 ) = (9)((1 / 3) x (1 / 3) - 1 ) = 3 x - 2 / 3 . (Rules 2 and 3) 13. f ( x ) = (3 x (1 / 2) ) = (3)( x 1 / 2 ) = (3)((1 / 2) x (1 / 2) - 1 ) = (3 / 2) x - 1 / 2 . (Rules 2 and 3) 15. f ( x ) = (7)( x - 12 ) = (7)( - 12 x - 12 - 1 ) = - 84 x - 13 . (Rules 2 and 3) 17. f ( x ) = (5 x 2 ) +( - 3 x ) +(7) = (5)( x 2 ) +( - 3)( x ) +0 = (5)(2 x 2 - 1 )+( - 3)( x 1 - 1 ) = 10 x - 3 . (Rules 1, 2, 3, and 4) 19. f ( x ) = ( - x 3 ) + (2 x 2 ) + ( - 6) = ( - 1)( x 3 ) + (2)( x 2 ) + 0 = ( - 1)(3 x 3 - 1 ) + (2)(2 x 2 - 1 ) = - 3 x 2 + 4 x. (Rules 1, 2, 3, and 4) 21. f ( x ) = (0 . 03 x 2 ) + ( - 0 . 4 x ) + (10) = (0 . 03)( x 2 ) + ( - 0 . 4)( x 1 ) + 0 = (0 . 03)(2 x 2 - 1 ) + ( - 0 . 4)( x 1 - 1 ) = 0 . 06 x - 0 . 4 . (Rules 1, 2, 3, and 4) 23. f ( x ) = ( x 3 x ) + ( - 4 x 2 x ) + ( 3 x ) = ( x 2 ) + ( - 4 x ) + (3 x - 1 ) = 2 x 2 - 1 + ( - 4)( x 1 ) + (3)( x - 1 ) = 2 x + ( - 4)( x 1 - 1 ) + (3)(( - 1) x - 1 - 1 ) = 2 x - 4 - 3 x - 2 . (Rules 2, 3, and 4) 25. f ( x ) = (4 x 4 ) + ( - 3 x (5 / 2) ) + (2) = (4)( x 4 ) + ( - 3)( x (5 / 2) ) + 0 = (4)(4 x 4 - 1 ) + ( - 3)((5 / 2) x (5 / 2) - 1 ) = 16 x 3 - (15 / 2) x 3 / 2 . (Rules 1, 2, 3, and 4) 27. f ( x ) = (3 x - 1 ) + (4 x - 2 ) = (3)( x - 1 ) + (4)( x - 2 ) = (3)(( - 1) x - 1 - 1 ) + (4)(( - 2) x - 2 - 1 ) = - 3 x - 2 - 8 x - 3 . (Rules 2, 3, and 4)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
29. f ( t ) = ( 4 t 4 ) + ( - 3 t 3 ) + ( 2 t ) = (4 t - 4 ) + ( - 3 t - 3 ) + (2 t - 1 ) = (4)( t - 4 ) + ( - 3)( t - 3 ) + (2)( t - 1 ) = (4)( - 4 t - 4 - 1 ) + ( - 3)( - 3 t - 3 - 1 ) + (2)( - t - 1 - 1 ) = - 16 t - 5 + 9 t - 4 - 2 t - 2 . (Rules 2, 3, and 4) 31. f ( x ) = (2 x ) + ( - 5 x ) = (2 x ) + ( - 5 x 1 / 2 ) = (2)( x ) + ( - 5)( x 1 / 2 ) = (2)( x 1 - 1 ) + ( - 5)((1 / 2) x 1 / 2 - 1 ) = 2 - (5 / 2) x - 1 / 2 . (Rules 2, 3, and 4) 33. f ( x ) = ( 2 x 2 ) + ( - 3 x 1 / 3 ) = (2 x - 2 ) + ( - 3 x - 1 / 3 ) = (2)( x - 2 ) + ( - 3)( x - 1 / 3 ) = (2)( - 2 x - 2 - 1 ) + ( - 3)(( - 1 / 3) x ( - 1 / 3) - 1 ) = - 4 x - 3 + x - 4 / 3 . (Rules 2, 3, and 4) 35. f ( x ) = (2 x 3 ) + ( - 4 x ) = (2)( x 3 ) + ( - 4)( x ) = (2)(3 x 3 - 1 ) + ( - 4)( x 1 - 1 ) = 6 x 2 - 4 . f ( - 2) = 6( - 2) 2 - 4 = 20 f (0) = 6(0) 2 - 4 = - 4 f ( - 2) = 6(2) 2 - 4 = 20 . 41. f ( x ) = (2 x 2 ) + ( - 3 x ) + (4) = (2)( x 2 ) + ( - 3)( x ) + 0 = (2)(2 x 2 - 1 ) + ( - 3)( x 1 - 1 ) = 4 x - 3 . The slope of the tangent line at the point (2,6) is f (2) = 4(2) - 3 = 5 . Thus the equation of the tangent line in slope intercept form is given by y = f (2) x + b = 5 x + b. To determine the value of b we substitute the x and y values at the point (2,6) into the equation and obtain 6 = 5(2) + b = 10 + b so that b = - 4 and the equation of the tangent line is y = 5 x - 4 . 43. f ( x ) = ( x 4 ) + ( - 3 x 3 ) + (2 x 2 ) + ( - x ) + (1) = ( x 4 ) + ( - 3)( x 3 ) + (2)( x 2 ) + ( - 1)( x ) + (1) = (4 x 4 - 1 ) + ( - 3)(3 x 3 - 1 ) + (2)(2 x 2 - 1 ) + ( - 1)( x 1 - 1 ) + 0 = 4 x 3 - 9 x 2 + 4 x - 1 .
The slope of the tangent line at the point (1,0) is f (1) = 4(1) 3 - 9(1) 2 + 4(1) - 1 = - 2 . Thus the equation of the tangent line in slope intercept form is given by y = f (1) x + b = - 2 x + b. To determine the value of b we substitute the x and y values at the point (1,0) into the equation and obtain 0 = - 2(1) + b = - 2 + b so that b = 2 the equation of the tangent line is y = - 2 x + 2 . 45. f ( x ) = 3 x 3 - 2 = 3 x 2 . The tangent line is horizontal when the slope is equal to zero, thus f ( x ) = 3 x 2 = 0 yields x = 0 and the corresponding point on the curve is (0 , f (0)) = (0 , 0) . 47. f ( x ) = ( x 3 ) + (1) = 3 x 3 - 2 + 0 = 3 x 2 . Since the slope of the tangent line at a point is equal to the value of the derivative at that point we write the equation f ( x ) = 3 x 2 = 12 and solving find x = ± 2; thus the corresponding points on the curve at which the tangent line has a slope of 12 are (2 , f (2)) = (2 , 9) and ( - 2 , f ( - 2)) = ( - 2 , - 7) . At (2,9) the tangent line has the equation y = 12 x + b.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
To determine the value of b we substitute the x and y values at the point (2,9) into the equation and obtain 9 = 12(2) + b so that b = - 15 the equation of the tangent line is y = 12 x - 15 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 36

HW3 - Section 3.1 1 f(x = 0(Rule 1 3 f(x = 5x51 = 5x4(Rule...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online