What is the tangent of f(x)=4sqrt(x),\at x=2 ?

The tangent of f(x)=4sqrt(x),\at x=2 is y=sqrt(2)x+2sqrt(2)

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tangent of f(x)=4sqrt(x),\at x=2

tangent of

f (x) = 4 x, at x = 2

y = 2 x + 22

Solution steps

Find the tangent point:

(2, 42)

Find the slope of

f (x) = 4 x : \frac{df ( x )}{d x} = \frac{2}{x}

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 = 2

Find the line with slope m=

2

and passing through

(2, 42) : y = 2 x + 22

y = 2 x + 22

\int_{1}^{8} x ln (x) d x

x \to 0 lim (\frac{∣ x - 1 ∣}{x - 1})

f (x) = sin^{2} (3 x + 2)

(e^{2 x} + e^{- x})^{^{'}}

y = - 3 - 5 x^{2}

What is the tangent of f(x)=4sqrt(x),\at x=2 ?
The tangent of f(x)=4sqrt(x),\at x=2 is y=sqrt(2)x+2sqrt(2)