f"(x)=10, f'(x)=3 f(0)=1/3 find f(1)
Sorry. If f'(x) is a constant, f"(x) = 0
Fix the typo and try again.
I will assume you meant f'(0) = 3
in that case,
f"(x) = 10
f'(x) = 10x+c
3 = 10*0+c -----> c = 3
f'(x) = 10x+3
f(x) = 5x^2+3x+c
1/3 = 5*0+3*0+c
c = 1/3
f(x) = 5x^2 + 3x + 1/3
I figure you can now find f(1).
To find f(1), we'll first need to find the expression for f(x). We are given that the second derivative of f(x), denoted as f"(x), is equal to 10, the first derivative of f(x), denoted as f'(x), is equal to 3, and f(0) is equal to 1/3.
We can start by integrating f"(x) to find the expression for f'(x), and then integrate f'(x) to find the expression for f(x).
Integrating f"(x) = 10 with respect to x gives:
f'(x) = ∫10 dx
f'(x) = 10x + C1
To find the value of C1, we use the given information that f'(x) = 3 when x = 0:
3 = 10(0) + C1
C1 = 3
Therefore, f'(x) = 10x + 3.
Next, we'll integrate f'(x) to find f(x):
f(x) = ∫(10x + 3) dx
f(x) = 5x^2 + 3x + C2
To find the value of C2, we use the given information that f(0) = 1/3:
1/3 = 5(0)^2 + 3(0) + C2
C2 = 1/3
Therefore, f(x) = 5x^2 + 3x + 1/3.
Finally, we can plug in x = 1 into the expression for f(x) to find f(1):
f(1) = 5(1)^2 + 3(1) + 1/3
f(1) = 5 + 3 + 1/3
f(1) = 8 + 1/3
f(1) = 25/3
Therefore, f(1) is equal to 25/3.