What is the tangent of f(x)=sqrt(x+1),(8,3) ?

The tangent of f(x)=sqrt(x+1),(8,3) is y= 1/6 x+5/3

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tangent of f(x)=sqrt(x+1),(8,3)

tangent of

f (x) = x + 1, (8, 3)

y = \frac{1}{6} x + \frac{5}{3}

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(8, 3) : y = \frac{1}{6} x + \frac{5}{3}

y = \frac{1}{6} x + \frac{5}{3}

\int \frac{4 x ^{2} - 64}{x} d x

y = ∣ x - 3 ∣, y = \frac{x}{2}, x = 0

\frac{d}{d x} (ln (\frac{e ^{7 x}}{1 + e ^{7 x}}))

3 tan^{2} (2 x)

\int_{1}^{2} \frac{3}{x ^{3} + 4 x} d x

What is the tangent of f(x)=sqrt(x+1),(8,3) ?
The tangent of f(x)=sqrt(x+1),(8,3) is y= 1/6 x+5/3