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

The tangent of f(x)=sqrt(x+3),\at x=1 is y= 1/4 x+7/4

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

tangent of

f (x) = x + 3, at x = 1

y = \frac{1}{4} x + \frac{7}{4}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = x + 3 : \frac{df ( x )}{d x} = \frac{1}{2 x + 3}

Compute the slope of

f (x) = x + 3

x = 1 : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(1, 2) : y = \frac{1}{4} x + \frac{7}{4}

y = \frac{1}{4} x + \frac{7}{4}

\frac{d}{d x} (\frac{ln ( x ^{5} )}{ln ( x ^{4} )})

\int_{- 2}^{2} (8 - 2 y^{2}) d y

x \to 0 lim (x^{2} + x - x)

\int 3 y 7 - 3 y^{2} d y

\int_{0}^{1} (y^{2} + y^{4}) d y

What is the tangent of f(x)=sqrt(x+3),\at x=1 ?
The tangent of f(x)=sqrt(x+3),\at x=1 is y= 1/4 x+7/4