What is the tangent of f(x)=(8x-7)^{1/2},\at x=2 ?

The tangent of f(x)=(8x-7)^{1/2},\at x=2 is y= 4/3 x+1/3

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tangent of f(x)=(8x-7)^{1/2},\at x=2

tangent of

f (x) = (8 x - 7)^{\frac{1}{2}}, at x = 2

y = \frac{4}{3} x + \frac{1}{3}

Solution steps

Find the tangent point:

(2, 3)

Find the slope of

f (x) = (8 x - 7)^{\frac{1}{2}} : \frac{df ( x )}{d x} = \frac{4}{( 8 x - 7 ) ^{\frac{1}{2}}}

Compute the slope of

f (x) = (8 x - 7)^{\frac{1}{2}}

x = 2 : m = \frac{4}{3}

Find the line with slope m=

\frac{4}{3}

and passing through

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

y = \frac{4}{3} x + \frac{1}{3}

n \to \infty lim (e^{- \frac{7}{n}})

f (x) = 2 x^{2} e^{x} + 4 x^{3}

\int 16 - x^{2} d x

\int [v + \frac{1}{( 3 v - 1 ) ^{3}}] d v

\int 2 \cdot x \cdot y

What is the tangent of f(x)=(8x-7)^{1/2},\at x=2 ?
The tangent of f(x)=(8x-7)^{1/2},\at x=2 is y= 4/3 x+1/3