What is the tangent of f(x)=(7x)/(2^x),\at x=2 ?

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

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tangent of f(x)=(7x)/(2^x),\at x=2

tangent of

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

y = \frac{7 ( 1 - 2 ln ( 2 ) )}{4} x + 7 ln (2)

Solution steps

Find the tangent point:

(2, \frac{7}{2})

Find the slope of

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

Compute the slope of

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

x = 2 : m = \frac{7 ( 1 - 2 ln ( 2 ) )}{4}

Find the line with slope m=

\frac{7 ( 1 - 2 ln ( 2 ) )}{4}

and passing through

(2, \frac{7}{2}) : y = \frac{7 ( 1 - 2 ln ( 2 ) )}{4} x + 7 ln (2)

y = \frac{7 ( 1 - 2 ln ( 2 ) )}{4} x + 7 ln (2)

\frac{d}{d x} ((x - 1) ln (x))

x \to 2 lim (x^{3} - 4)

a sin (\frac{( π \cdot x )}{b})

\frac{d}{d x} (\frac{64}{x ^{2} + 16})

\int \frac{x ^{2} - a ^{2}}{x ^{2}} d x

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