I really dont understand how to start with this?

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

First calculate the second derivative,

Then write down the quadratic approximation,

That yields,

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

Hope this helps.

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

Step-by-step 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: -0.5 would be the most likley outcome

step-by-step explanation: