Find f(x) satisfying the given conditions:
1)
f"(x) = 1/x^3
f'(1) = 1/2
f(1) = 0
2)
f"'(x) = sin (x)
f"(0) = 0
f'(0) = 1
f(0) = 0
1 )
f" ( x ) = 1 / x ^ 3 = x ^ ( -3 )
f´ ( x ) = integral of x ^ ( -3 ) dx
f´ ( x ) = x ^ ( - 3 + 1 ) / ( - 3 + 1 ) + C
f´ ( x ) = x ^ ( - 2 ) / ( - 2 ) + C
f´ ( x ) = 1 / x ^ 2 / ( - 2 ) + C
f´ ( x ) = - 1 / ( 2 x ^ 2 ) + C
f´ ( 1 ) = 1 / 2
- 1 / ( 2 * 1 ^ 2 ) + C = 1 / 2
- 1 / ( 2 * 1 ) + C = 1 / 2
- 1 / 2 + C = 1 / 2 Add 1 / 2 to both sides
- 1 / 2 + C + 1 / 2 = 1 / 2 + 1 / 2
C = 1
f´ ( x ) = - 1 / ( 2 x ^ 2 ) + C
f´ ( x ) = - 1 / ( 2 x ^ 2 ) + 1
f ( x ) = integral of f´ ( x ) dx
f ( x ) = integral of [ - 1 / ( 2 x ^ 2 ) ] dx + integral of dx
f ( x ) = - 1 / 2 [ integral of ( 1 / x ^ 2 ) ] dx + integral of dx
f ( x ) = - 1 / 2 [ integral of x ^ ( - 2 ) ] + integral of dx
f ( x ) = - 1 / 2 [ x ^ ( - 2 + 1 ) / ( - 2 + 1 ) ] + integral of dx
f ( x ) = - 1 / 2 [ x ^ ( - 1 ) / ( - 1 ) ] + x + C1
f ( x ) = - 1 / 2 [ - ( 1 / x ) ] + x + C1
f ( x ) = 1 / 2 [ ( 1 / x ) ] + x + C1
f ( x ) = x + 1 / 2 x + C1
f ( 1 ) = 0
1 + 1 / ( 2 * 1 ) + C1 = 0
1 + 1 / 2 + C1 = 0
2 / 2 + 1 / 2 + C1 = 0
3 / 2 + C1 = 0 Subtract 3 / 2 to both sides
3 / 2 + C1 - 3 / 2 = 0 - 3 / 2
C1 = - 3 / 2
So :
f ( x ) = x + 1 / 2 x + C 1
f ( x ) = x + 1 / 2 x - 3 / 2
2 )
f " ( x ) = integral of sin (x) dx
f " ( x ) = - cos ( x ) + C
f " ( 0 ) = 0
- cos ( 0 ) + C = 0
- 1 + C = 0 Add 1 to both sides
- 1 + C + 1 = 0 + 1
C = 1
f " ( x ) = - cos ( x ) + C
f " ( x ) = - cos ( x ) + 1
f´( x ) = integral of f " ( x ) dx
f´( x ) = integral of [ - cos ( x ) + 1 ] dx
f´( x ) = integral of [ - cos ( x ) ] dx + integral of dx
f´( x ) = - integral of [ cos ( x ) ] dx + integral of dx
f´( x ) = - sin ( x ) + x + C1
f´( x ) = x - sin ( x ) + C1
f´( 0 ) = 1
0 - sin ( 0 ) + C1 = 1
0 - 0 + C1 = 1
C1 = 1
f´( x ) = x - sin ( x ) + 1
f ( x ) = integral of f´( x ) dx
f ( x ) = integral of [ x - sin ( x ) + 1 ] dx
f ( x ) = integral of x dx - integral of sin ( x ) + integral of dx
f ( x ) = x ^ 2 / 2 - [ - cos ( x ) ] + x + C2
f ( x ) = x ^ 2 / 2 + cos ( x ) + x + C2
f ( x ) = x ^ 2 / 2 + x + cos ( x ) + C2
f ( 0 ) = 0
0 ^ 2 / 2 + cos ( 0 ) + 0 + C2 = 0
0 + 1 + 0 + C2 = 0
1 + C2 = 0 Subtract 1 to both sides
1 + C2 - 1 = 0 - 1
C2 = - 1
So :
f ( x ) = x ^ 2 / 2 + x + cos ( x ) + C2
f ( x ) = x ^ 2 / 2 + x + cos ( x ) - 1
To find the function f(x) that satisfies the given conditions, we will first integrate the derivatives to obtain the antiderivatives, and then use the initial conditions to determine the specific function.
1) For the first condition:
Given: f"(x) = 1/x^3
Integrating both sides once will give us:
f'(x) = ∫ (1/x^3) dx
To compute this integral, we can rewrite the expression as:
f'(x) = x^-3 dx
Now, using the power rule of integration, we can integrate x^-3:
f'(x) = (-1/2) * x^-2 + C1
where C1 is the constant of integration.
Given: f'(1) = 1/2
Plugging in x = 1 and f'(1) = 1/2 into the equation above, we can solve for C1:
1/2 = (-1/2) * 1^-2 + C1
1/2 = (-1/2) + C1
1/2 + 1/2 = C1
C1 = 1
So, f'(x) = (-1/2) * x^-2 + 1.
Next, we integrate f'(x) to obtain f(x):
f(x) = ∫((-1/2) * x^-2 + 1) dx
Integrating -1/2 * x^-2 gives:
f(x) = (1/2) * (-1) * x^-1 + x + C2
where C2 is another constant of integration.
Given: f(1) = 0
Plugging in x = 1 and f(1) = 0 into the equation above, we can solve for C2:
0 = (1/2) * (-1) * 1^-1 + 1 + C2
0 = -1/2 + 1 + C2
1/2 = C2
So, f(x) = (1/2) * (-1) * x^-1 + x + (1/2).
Therefore, the function f(x) satisfying the given conditions is:
f(x) = -1/(2x) + x + 1/2.
2) For the second condition:
Given: f"'(x) = sin(x)
Integrating both sides once will give us:
f"(x) = ∫ sin(x) dx
The integral of sin(x) is -cos(x), so:
f"(x) = -cos(x) + C1
where C1 is the constant of integration.
Given: f"(0) = 0
Plugging in x = 0 and f"(0) = 0 into the equation above, we can solve for C1:
0 = -cos(0) + C1
0 = -1 + C1
C1 = 1
So, f"(x) = -cos(x) + 1.
Next, we integrate f"(x) to obtain f'(x):
f'(x) = ∫(-cos(x) + 1) dx
Integrating -cos(x) gives:
f'(x) = sin(x) + x + C2
where C2 is another constant of integration.
Given: f'(0) = 1
Plugging in x = 0 and f'(0) = 1 into the equation above, we can solve for C2:
1 = sin(0) + 0 + C2
1 = 0 + C2
C2 = 1
So, f'(x) = sin(x) + x + 1.
Finally, we integrate f'(x) to obtain f(x):
f(x) = ∫(sin(x) + x + 1) dx
Integrating sin(x) and x gives:
f(x) = -cos(x) + (1/2) * x^2 + x + C3
where C3 is another constant of integration.
Given: f(0) = 0
Plugging in x = 0 and f(0) = 0 into the equation above, we can solve for C3:
0 = -cos(0) + (1/2) * 0^2 + 0 + C3
0 = -1 + C3
C3 = 1
So, f(x) = -cos(x) + (1/2) * x^2 + x + 1.
Therefore, the function f(x) satisfying the given conditions is:
f(x) = -cos(x) + (1/2) * x^2 + x + 1.