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

The tangent of f(x)=sqrt(x+1),(3,2) is y= 1/4 x+5/4

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tangent of f(x)=sqrt(x+1),(3,2)

tangent of

f (x) = x + 1, (3, 2)

y = \frac{1}{4} x + \frac{5}{4}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x + 1

(3, 2) : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(3, 2) : y = \frac{1}{4} x + \frac{5}{4}

y = \frac{1}{4} x + \frac{5}{4}

\int C e^{- a t} d t

f (x) = \frac{6}{x ^{4}}

y^{^{'}} = \frac{x + 2 y}{x}

\int e^{- l n (x)} d x

What is the tangent of f(x)=sqrt(x+1),(3,2) ?
The tangent of f(x)=sqrt(x+1),(3,2) is y= 1/4 x+5/4