What is the tangent of f(x)=8sqrt(x),(4,16) ?

The tangent of f(x)=8sqrt(x),(4,16) is y=2x+8

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tangent of f(x)=8sqrt(x),(4,16)

tangent of

f (x) = 8 x, (4, 16)

y = 2 x + 8

Solution steps

Find the slope of

f (x) = 8 x : \frac{df ( x )}{d x} = \frac{4}{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

(4, 16) : y = 2 x + 8

y = 2 x + 8

y = 3 x^{2}, y^{2} = \frac{1}{3} x

\int \frac{ln ^{2} ( x ^{2} )}{x} d x

\frac{\partial}{\partial x} (y e^{y} x)

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

f (X) = 7 (X - 5)^{\frac{2}{3}}

What is the tangent of f(x)=8sqrt(x),(4,16) ?
The tangent of f(x)=8sqrt(x),(4,16) is y=2x+8