Question

let f be a differentiable function such that f(1)=π and f'(x)=√x^3+6. what is the value of f(5)?

Answer 1

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) = $\sqrt{x^{3}}$ + 6

- Change the root to fraction power

∵ $\sqrt{x^{3}}$ = $x^{\frac{3}{2}}$

∴ f'(x) = $x^{\frac{3}{2}}$ + 6

∴ f(x) = ∫ $x^{\frac{3}{2}}$ + 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) = $\frac{x^{\frac{5}{2}}}{\frac{5}{2}}$ + 6x + c

- c is the constant of integration

∵ $\frac{x^{\frac{5}{2}}}{\frac{5}{2}}=\frac{2}{5}x^{\frac{5}{2}}$

∴ f(x) = $\frac{2}{5}$ $x^{\frac{5}{2}}$ + 6x + c

- To find c substitute x by 1 and f(x) by π

∴ π = $\frac{2}{5}$ $(1)^{\frac{5}{2}}$ + 6(1) + c

∴ π = $\frac{2}{5}$ + 6 + c

∴ π = 6.4 + c

- Subtract 6.4 from both sides

∴ c = - 3.2584

∴ f(x) = $\frac{2}{5}$ $x^{\frac{5}{2}}$ + 6x - 3.2584

To find f(5) Substitute x by 5

∵ x = 5

∴ f(5) = $\frac{2}{5}$ $(5)^{\frac{5}{2}}$ + 6(5) - 3.2584

∴ f(5) = 49.1

The value of f(5) is 49.1

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Answer 2

By integrating f'(x) and knowing the value of f(1), we will get f(5) = 49.1

So we know that:

$f'(x) = \sqrt{x^3} + 6$

Integrating that, we get:

$f(x) = \int\limits {\sqrt{x^3} + 6} \, dx = \frac{2}{5}*x^{5/2} + 6*x + c$

Where c is a constant of integration, to get its value, we need to use the condition:

f(1) = π

Then we have:

$f(1) = 3.14 = \frac{2}{5}*1^{5/2} + 6*1 + c\\\\3.14 = \frac{32}{5} + c\\\\3.14 - 6.4 = c = -3.26$

Then:

$f(x) = \frac{2}{5}*x^{5/2} + 6*x -3.26$

Now we just want to evaluate this in x = 5:

$f(5) = \frac{2}{5}*5^{5/2} + 6*5 -3.26 = 49.1$

So the value of f(5) is 49.1

If you want to learn more, you can read:

https://brainly.com/question/18125359