What is the tangent of f(x)=(64)/(x^2+16),\at x=-4 ?

The tangent of f(x)=(64)/(x^2+16),\at x=-4 is y= 1/2 x+4

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tangent of f(x)=(64)/(x^2+16),\at x=-4

tangent of

f (x) = \frac{64}{x ^{2} + 16}, at x = - 4

y = \frac{1}{2} x + 4

Solution steps

Find the tangent point:

(- 4, 2)

Find the slope of

f (x) = \frac{64}{x ^{2} + 16} : \frac{df ( x )}{d x} = - \frac{128 x}{( x ^{2} + 16 ) ^{2}}

Compute the slope of

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

x = - 4 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

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

y = \frac{1}{2} x + 4

5 e^{- t}

\frac{d}{d x} (arcsin (x))

x \to - 1 lim (1 - x^{2})

y = x^{2} - 2 x, at x = 0

\int \frac{ln ( x )}{x 1 + 7 ln ^{2} ( x )} d x

What is the tangent of f(x)=(64)/(x^2+16),\at x=-4 ?
The tangent of f(x)=(64)/(x^2+16),\at x=-4 is y= 1/2 x+4