What is the tangent of f(x)=x^2+4/x-10,\at x=8 ?

The tangent of f(x)=x^2+4/x-10,\at x=8 is y= 255/16 x-73

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tangent of f(x)=x^2+4/x-10,\at x=8

tangent of

f (x) = x^{2} + \frac{4}{x} - 10, at x = 8

y = \frac{255}{16} x - 73

Solution steps

Find the tangent point:

(8, \frac{109}{2})

Find the slope of

f (x) = x^{2} + \frac{4}{x} - 10 : \frac{df ( x )}{d x} = 2 x - \frac{4}{x ^{2}}

Compute the slope of

f (x) = x^{2} + \frac{4}{x} - 10

x = 8 : m = \frac{255}{16}

Find the line with slope m=

\frac{255}{16}

and passing through

(8, \frac{109}{2}) : y = \frac{255}{16} x - 73

y = \frac{255}{16} x - 73

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What is the tangent of f(x)=x^2+4/x-10,\at x=8 ?
The tangent of f(x)=x^2+4/x-10,\at x=8 is y= 255/16 x-73