What is the tangent of f(x)=sqrt(x^2+3),\at x= 1/2 ?

The tangent of f(x)=sqrt(x^2+3),\at x= 1/2 is y= 1/(sqrt(13))x+6/(sqrt(13))

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tangent of f(x)=sqrt(x^2+3),\at x= 1/2

tangent of

f (x) = x^{2} + 3, at x = \frac{1}{2}

y = \frac{1}{13} x + \frac{6}{13}

Solution steps

Find the tangent point:

(\frac{1}{2}, \frac{13}{2})

Find the slope of

f (x) = x^{2} + 3 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 3}

Compute the slope of

f (x) = x^{2} + 3

x = \frac{1}{2} : m = \frac{1}{13}

Find the line with slope m=

\frac{1}{13}

and passing through

(\frac{1}{2}, \frac{13}{2}) : y = \frac{1}{13} x + \frac{6}{13}

y = \frac{1}{13} x + \frac{6}{13}

\frac{\partial}{\partial y} (cos (x z))

\int (3 t e^{3 t}) d t

\int_{- 2}^{1} x^{2} - 1 d x

\frac{\partial}{\partial x} (arcsin (2 x - 5))

x = y^{2}, x = 36

What is the tangent of f(x)=sqrt(x^2+3),\at x= 1/2 ?
The tangent of f(x)=sqrt(x^2+3),\at x= 1/2 is y= 1/(sqrt(13))x+6/(sqrt(13))