What is the tangent of f(x)=sqrt(2x+5),\at x=2 ?

The tangent of f(x)=sqrt(2x+5),\at x=2 is y= 1/3 x+7/3

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tangent of f(x)=sqrt(2x+5),\at x=2

tangent of

f (x) = 2 x + 5, at x = 2

y = \frac{1}{3} x + \frac{7}{3}

Solution steps

Find the tangent point:

(2, 3)

Find the slope of

f (x) = 2 x + 5 : \frac{df ( x )}{d x} = \frac{1}{2 x + 5}

Compute the slope of

f (x) = 2 x + 5

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

Find the line with slope m=

\frac{1}{3}

and passing through

(2, 3) : y = \frac{1}{3} x + \frac{7}{3}

y = \frac{1}{3} x + \frac{7}{3}

n = 0 \sum \infty \frac{10 0 ^{n}}{n !}

\frac{1 - 2 s}{s ^{2} + 4 s + 5}

\frac{d}{d x} (2 sin (8 x) cos (x))

\frac{d}{d x} ((\frac{x ^{3}}{x ^{4} - 2 x})^{3})

\int (2 x + 1)^{\frac{1}{2}} d x

What is the tangent of f(x)=sqrt(2x+5),\at x=2 ?
The tangent of f(x)=sqrt(2x+5),\at x=2 is y= 1/3 x+7/3