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

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

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

tangent of

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

y = \frac{3}{2} x + \frac{1}{2}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

f (x) = x + x

x = 1 : m = \frac{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

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

y = \frac{3}{2} x + \frac{1}{2}

\int_{0}^{1} 2^{- θ} d θ

x \to 0 lim (3 x - 4)

\frac{d}{d y} (cos (x y))

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 6)^{4})

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