What is the tangent of f(x)=8sqrt(x)-3,\at x=64 ?

The tangent of f(x)=8sqrt(x)-3,\at x=64 is y= 1/2 x+29

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tangent of f(x)=8sqrt(x)-3,\at x=64

tangent of

f (x) = 8 x - 3, at x = 64

y = \frac{1}{2} x + 29

Solution steps

Find the tangent point:

(64, 61)

Find the slope of

f (x) = 8 x - 3 : \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 = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(64, 61) : y = \frac{1}{2} x + 29

y = \frac{1}{2} x + 29

\frac{d}{d y} (y x^{2})

y^{^{''}} + 11 y^{^{'}} + 22 y = 3468 e^{2 t} cos (12 t)

\int_{0}^{1} π (e^{x} + 1)^{2} d x

2 y \frac{d y}{d x} + y^{2} cot (x) = csc (x)

x \to 0 lim ((2 x - 2)^{3})

What is the tangent of f(x)=8sqrt(x)-3,\at x=64 ?
The tangent of f(x)=8sqrt(x)-3,\at x=64 is y= 1/2 x+29