What is the tangent of f(x)= x/(x^2-15),\at x=4 ?

The tangent of f(x)= x/(x^2-15),\at x=4 is y=-31x+128

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tangent of f(x)= x/(x^2-15),\at x=4

tangent of

f (x) = \frac{x}{x ^{2} - 15}, at x = 4

y = - 31 x + 128

Solution steps

Find the tangent point:

(4, 4)

Find the slope of

f (x) = \frac{x}{x ^{2} - 15} : \frac{df ( x )}{d x} = \frac{- x ^{2} - 15}{( x ^{2} - 15 ) ^{2}}

Compute the slope of

f (x) = \frac{x}{x ^{2} - 15}

x = 4 : m = - 31

Find the line with slope m=

- 31

and passing through

(4, 4) : y = - 31 x + 128

y = - 31 x + 128

y^{^{''}} + 25 y^{^{'}} = 0

f (x) = 7 e^{x^{2} - 5 x + 6}

\int (10 x + 4) 5 x^{2} + 4 x d x

\int_{1}^{3} ((a x)^{3} + e^{- b x}) d x

\int_{0}^{2} 4 x d x

What is the tangent of f(x)= x/(x^2-15),\at x=4 ?
The tangent of f(x)= x/(x^2-15),\at x=4 is y=-31x+128