Question

I really dont understand how to start with this?​

Answer 1

f(x) = x² + (10/3)(x^1.5) - 6/x - 10sqrt(3)

Explanation:

f'(x) is the derivative of f(x)

To go back to f(x), integrate f'(x)

f'(x) = 2x + 5(x^0.5) + 6(x^-2)

Add 1 to the power, divide by the new power

2(x^2)/2 + 5(x^1.5)/1.5 + 6(x^-1)/-1

f(x) = x² + (10/3)(x^1.5) - 6/x + c

Find c using (3,7)

7 = 3² + (10/3)(3^1.5) -6/3 + c

7 = 9 + 10sqrt(3) - 2 + c

c = - 10sqrt(3)

f(x) = x² + (10/3)(x^1.5) - 6/x - 10sqrt(3)

Answer 2

You could determine the quadratic approximation of the curve near
`x=3` which gives you a good estimate since the function is relatively monotone.

First calculate the second derivative,

`f''(x)=\Big(f'(x)\Big)'=\Big(2x+5√(x)+(6)/(x^2)\Big)'=2+(5)/(2√(x))-(12)/(x^3)`

Then write down the quadratic approximation,

`f(x)\approx f(3)+f'(3)x+(f''(3))/(2)\Big(x-3\Big)^2`

That yields,

`f(x)\approx\boxed{9+6(2)/(3)x+5√(3)x+\Big((5)/(2√(3))-(4)/(9)\Big)\Big(x-3\Big)^2}`

Of course this can be further simplified but its not obligatory.

Hope this helps.