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F’(x)= 1+3 square root x f(9)=68

Sagot :

Averysmartperson

Answer:

[tex]f(x) = x + 2x^\frac{3}{2} + 5[/tex]

Step-by-step explanation:

I'm assuming the question here is asking for f(x)

The first step would be to find the antiderivative of f'(x)

[tex]\int{1+3\sqrt{x}} \, dx = \int{1 + 3x^\frac{1}{2}} dx[/tex]

Apply the reverse power rule

[tex]x + 2x^\frac{3}{2} + C[/tex]

Now solve for C when x = 9

[tex]x + 2x^\frac{3}{2} + C = 68\\9 + 2(9)^\frac{3}{2} + C = 68\\9 + 2(27) + C = 68\\9 + 54 + C = 68\\63 + C = 68\\C = 5[/tex]

Now substitute into the antiderivative

[tex]f(x) = x + 2x^\frac{3}{2} + 5[/tex]

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