What is the tangent of f(x)=4sqrt(x)-1,\at x=16 ?

The tangent of f(x)=4sqrt(x)-1,\at x=16 is y= 1/2 x+7

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tangent of f(x)=4sqrt(x)-1,\at x=16

tangent of

f (x) = 4 x - 1, at x = 16

y = \frac{1}{2} x + 7

Solution steps

Find the tangent point:

(16, 15)

Find the slope of

f (x) = 4 x - 1 : \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 = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(16, 15) : y = \frac{1}{2} x + 7

y = \frac{1}{2} x + 7

y = 3 x, y = x^{2} - 4

y^{^{'}} = x + 6 y

\frac{d}{d x} (ln (\frac{x}{3}) + \frac{x - 3}{x})

\frac{3}{s ( s - 1 ) ^{3}}

f (x) = lo g_{10} (1 - x)

What is the tangent of f(x)=4sqrt(x)-1,\at x=16 ?
The tangent of f(x)=4sqrt(x)-1,\at x=16 is y= 1/2 x+7