Answer :

r3t40

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

First calculate the second derivative,

[tex]f''(x)=\Big(f'(x)\Big)'=\Big(2x+5\sqrt{x}+\dfrac{6}{x^2}\Big)'=2+\dfrac{5}{2\sqrt{x}}-\dfrac{12}{x^3}[/tex]

Then write down the quadratic approximation,

[tex]f(x)\approx f(3)+f'(3)x+\dfrac{f''(3)}{2}\Big(x-3\Big)^2[/tex]

That yields,

[tex]f(x)\approx\boxed{9+6\dfrac{2}{3}x+5\sqrt{3}x+\Big(\dfrac{5}{2\sqrt{3}}-\dfrac{4}{9}\Big)\Big(x-3\Big)^2}[/tex]

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

Hope this helps.

amna04352

Answer:

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)

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