What is the tangent of f(x)=(12x-3)^{1/2},\at x=1 ?

The tangent of f(x)=(12x-3)^{1/2},\at x=1 is y=2x+1

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tangent of f(x)=(12x-3)^{1/2},\at x=1

tangent of

f (x) = (12 x - 3)^{\frac{1}{2}}, at x = 1

y = 2 x + 1

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = (12 x - 3)^{\frac{1}{2}} : \frac{df ( x )}{d x} = \frac{6}{( 12 x - 3 ) ^{\frac{1}{2}}}

Compute the slope of

f (x) = (12 x - 3)^{\frac{1}{2}}

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 3) : y = 2 x + 1

y = 2 x + 1

x \to - \infty lim (e^{x} - x)

\frac{d}{d x} (cot^{2} (5 x))

\frac{d}{d x} (cos (3 sin (x)))

\frac{d ^{2}}{d x ^{2}} y + 49 y = 0

\frac{d}{d x} ((3 x - 1)^{2})

What is the tangent of f(x)=(12x-3)^{1/2},\at x=1 ?
The tangent of f(x)=(12x-3)^{1/2},\at x=1 is y=2x+1