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

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

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tangent of f(x)=sqrt(2x+1),(4,3)

tangent of

f (x) = 2 x + 1, (4, 3)

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

Solution steps

Find the slope of

f (x) = 2 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}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

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

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

(x^{3} - 4 x) (2 x^{2} + 5 x)

\int e^{t (a - s)} d t

\int_{0}^{1} cos^{2} (3 π x) d x

n = 1 \sum \infty n p (1 - p)^{n}

\frac{d}{d x} (x^{4} (1 - \frac{2}{x + 1}))

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