# g(x)=#int_sqrtx^xtan^(-1)tdt# Find? 1) g(#sqrt3#) 2) g'(#sqrt3#) 3) g"(#sqrt3#)

Please see below.

Here,

#:.g(sqrt3)=sqrt3xxpi/3-root(4)3tan^-1root(4)3-1/2ln| (1+sqrt3)/(1+root(4)3)|#

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To find the value of ( g(\sqrt{3}) ), we need to evaluate the definite integral ( \int_{\sqrt{3}}^{\sqrt{3}} t \tan^{-1}(t) , dt ). Since the upper and lower limits of integration are the same, the integral evaluates to zero. Therefore, ( g(\sqrt{3}) = 0 ).

To find the first derivative ( g'(\sqrt{3}) ), we can use the Fundamental Theorem of Calculus. Differentiating under the integral sign with respect to ( x ), we get ( g'(x) = \frac{d}{dx} \int_{\sqrt{x}}^{x} t \tan^{-1}(t) , dt ). Applying the Fundamental Theorem of Calculus, we get ( g'(x) = x \tan^{-1}(x) - \frac{1}{2} \ln(x^2 + 1) + \frac{\sqrt{x}}{2} ). Evaluating this at ( x = \sqrt{3} ), we find ( g'(\sqrt{3}) ).

To find the second derivative ( g''(\sqrt{3}) ), we differentiate ( g'(x) ) with respect to ( x ). After differentiation, we evaluate ( g''(\sqrt{3}) ) to find the second derivative at ( x = \sqrt{3} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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