Let f be a differentiable function such that f(1)=π and f'(x)=√x^3+6. what is the value of f(5)?
1 answer:
The value of f(5) is 49.1
Step-by-step explanation:
To find f(x) from f'(x) use the integration
f(x) = ∫ f'(x)
1. Find The integration of f'(x) with the constant term
2. Substitute x by 1 and f(x) by π to find the constant term
3. Write the differential function f(x) and substitute x by 5 to find f(5)
∵ f'(x) = + 6
- Change the root to fraction power
∵ =
∴ f'(x) = + 6
∴ f(x) = ∫ + 6
- In integration add the power by 1 and divide the coefficient by the
new power and insert x with the constant term
∴ f(x) = + 6x + c
- c is the constant of integration
∵
∴ f(x) =
+ 6x + c
- To find c substitute x by 1 and f(x) by π
∴ π =
+ 6(1) + c
∴ π = + 6 + c
∴ π = 6.4 + c
- Subtract 6.4 from both sides
∴ c = - 3.2584
∴ f(x) =
+ 6x - 3.2584
To find f(5) Substitute x by 5
∵ x = 5
∴ f(5) =
+ 6(5) - 3.2584
∴ f(5) = 49.1
The value of f(5) is 49.1
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