f'(x) = 1 +3x, f(4) 2532. f'(x) 5x3x4. f(1) 2f'(t) = 4/(1 + 12), f(1) 034. f'()=1 +1/1, 0, f(1)635. f'(x)5x2/, f(8)2136. f(x)=(x 1)/Vx, f1)= 5T/2 < 0,f(2) 3, f'(1) =43. f"(x)= 4 6x + 24x2, f(0)= 3, f(1)= 1044. f"(x)= x3 sinh x,f (0) = 1,f (2) 2.6ex- 2 sin x, f(0)= 3, f(T/ 2) = 045. f"(x)=t - cos t, f(0) = 2, f(1)= 246.f"(t)f(1) 23COs t,47. f"(x)= x-2, x> 0, f(1) = 0, f(2) = 048. f"(x)= coS x, f(0)= 1, f'(0)= 2, f"(0)49. Given that the graph of f passes through the poi(2, 5) and that the slope of its tangent line at (x,is 3 - 4x, find f(1).0. Find a function f such that f'(x)= x and thet y= 0 is tangent to the graph of f.11

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Asked Dec 3, 2019
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I need help with question #45 in Section 4.9, page 356, in the James Stewart Calculus Eighth Edition textbook. I need to find "f" given the function and the conditions.

f'(x) = 1 +3x, f(4) 25
32. f'(x) 5x
3x
4. f(1) 2
f'(t) = 4/(1 + 12), f(1) 0
34. f'()=1 +1/1, 0, f(1)6
35. f'(x)5x2/, f(8)
21
36. f(x)=(x 1)/Vx, f1)= 5
T/2 <<T/2,
37. f (t)= sec t (sec 1 tan t),
f(m/4) 1
38. f'(t) 3' 3/t, f(1)=2, f(-1) 1
39. f"(x)= -212x 12x2, f(0) 4, f'(0) 1
40. f"(x)= 8x3 5, f(1) = 0, f'(1)= 8
41. f"(0)s
f(0)= 3, f'(0)= 4
sin 0cos 0,
42. f"(t)t 1/t2, t> 0,
f(2) 3, f'(1) =
43. f"(x)= 4 6x + 24x2, f(0)= 3, f(1)= 10
44. f"(x)= x3 sinh x,
f (0) = 1,
f (2) 2.6
ex- 2 sin x, f(0)= 3, f(T/ 2) = 0
45. f"(x)=
t - cos t, f(0) = 2, f(1)= 2
46.f"(t)
f(1) 2
3
COs t,
47. f"(x)= x-2, x> 0, f(1) = 0, f(2) = 0
48. f"(x)= coS x, f(0)= 1, f'(0)= 2, f"(0)
49. Given that the graph of f passes through the poi
(2, 5) and that the slope of its tangent line at (x,
is 3 - 4x, find f(1).
0. Find a function f such that f'(x)= x and the
t y= 0 is tangent to the graph of f.
11
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f'(x) = 1 +3x, f(4) 25 32. f'(x) 5x 3x 4. f(1) 2 f'(t) = 4/(1 + 12), f(1) 0 34. f'()=1 +1/1, 0, f(1)6 35. f'(x)5x2/, f(8) 21 36. f(x)=(x 1)/Vx, f1)= 5 T/2 <<T/2, 37. f (t)= sec t (sec 1 tan t), f(m/4) 1 38. f'(t) 3' 3/t, f(1)=2, f(-1) 1 39. f"(x)= -212x 12x2, f(0) 4, f'(0) 1 40. f"(x)= 8x3 5, f(1) = 0, f'(1)= 8 41. f"(0)s f(0)= 3, f'(0)= 4 sin 0cos 0, 42. f"(t)t 1/t2, t> 0, f(2) 3, f'(1) = 43. f"(x)= 4 6x + 24x2, f(0)= 3, f(1)= 10 44. f"(x)= x3 sinh x, f (0) = 1, f (2) 2.6 ex- 2 sin x, f(0)= 3, f(T/ 2) = 0 45. f"(x)= t - cos t, f(0) = 2, f(1)= 2 46.f"(t) f(1) 2 3 COs t, 47. f"(x)= x-2, x> 0, f(1) = 0, f(2) = 0 48. f"(x)= coS x, f(0)= 1, f'(0)= 2, f"(0) 49. Given that the graph of f passes through the poi (2, 5) and that the slope of its tangent line at (x, is 3 - 4x, find f(1). 0. Find a function f such that f'(x)= x and the t y= 0 is tangent to the graph of f. 11

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Expert Answer

Step 1

Consider the given second order differential equation.

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f"(x)e-2 sinx

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Step 2

First of all, integrate f ''(x) w.r.t x.

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f"(x) =e* – 2sin .xr y"= e* – 2sin x y'= [(e* - 2 = fe'dx – 2[sin xdx sin x \d+ f '(x) = e* +2cos.x + C

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Step 3

Again, integrate f '(x) w.r...

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y'e 2cosr+C y = [(e* +2cosx+G) =fe'dx+ 2[cos.xdx + fc,dc f(x)e +2sin x+ Cx +C

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