What is the tangent of 5-4/(x^2),\at x=4 ?

The tangent of 5-4/(x^2),\at x=4 is y= 1/8 x+17/4

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tangent of 5-4/(x^2),\at x=4

tangent of

5 - \frac{4}{x ^{2}}, at x = 4

y = \frac{1}{8} x + \frac{17}{4}

Solution steps

Find the tangent point:

(4, \frac{19}{4})

Find the slope of

y = 5 - \frac{4}{x ^{2}} : \frac{d y}{d x} = \frac{8}{x ^{3}}

Compute the slope of

y = 5 - \frac{4}{x ^{2}}

x = 4 : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(4, \frac{19}{4}) : y = \frac{1}{8} x + \frac{17}{4}

y = \frac{1}{8} x + \frac{17}{4}

x \to \infty lim (x^{3})

\frac{2 s}{s ^{2} + 6 s + 9}

\int_{1}^{e^{8}} \frac{ln ^{2} ( x ^{2} )}{x} d x

4 x^{2} - x, 1, 3

\frac{\partial ^{2}}{\partial u ( v ) \partial v} (e^{u (v)} + cos (v))

What is the tangent of 5-4/(x^2),\at x=4 ?
The tangent of 5-4/(x^2),\at x=4 is y= 1/8 x+17/4