Question

Find f.f '(x) = square root( x) (6 + 5x), f(1) = 7

Answer 1

Since we have the first derivative of function f, then, we can find f as follows:

`f(x)=\int f^(\prime)(x)dx`

We can rewrite the given expression as follows

`f^(\prime)(x)=6x^{(1)/(2)}+5x^{(3)/(2)}`

because

`6√(x)=6x^{(1)/(2)}`

and

`5x√(x)=5x^{(3)/(2)}`

So, we need to compute

`f(x)=\int(6x^{(1)/(2)}+5x^{(3)/(2)})dx`

From the integration formula:

`\int x^ndx=(x^(n+1))/(n+1)`

we get

`f(x)=6\frac{x^{(1)/(2)+1}}{(1)/(2)+1}+5\frac{x^{(3)/(2)+1}}{(3)/(2)+1}+C`

where C is the constant of integration. From this result, we have

`f(x)=6\frac{x^{(3)/(2)}}{(3)/(2)}+5\frac{x^{(5)/(2)}}{(5)/(2)}+C`

which gives

`f(x)=4x^{(3)/(2)}+2x^{(5)/(2)}+C`

or equivalently,

`f(x)=4√(x^3)+2√(x^5)+C`

Finally, we can find C by substituting the given information about f(x), that is, f(1)=7. It yields,

`7=4√(1^3)+2√(1^5)+C`

which gives

`\begin{gathered} 7=4+2+C \\ 7=6+C \end{gathered}`

Then

`C=1`

Therefore, the answer is:

`f(x)=4√(x^3)+2√(x^5)+1`