What is the tangent of f(x)=(125)/(x^2+24),\at x=-1 ?

The tangent of f(x)=(125)/(x^2+24),\at x=-1 is y= 2/5 x+27/5

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tangent of f(x)=(125)/(x^2+24),\at x=-1

tangent of

f (x) = \frac{125}{x ^{2} + 24}, at x = - 1

y = \frac{2}{5} x + \frac{27}{5}

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

f (x) = \frac{125}{x ^{2} + 24} : \frac{df ( x )}{d x} = - \frac{250 x}{( x ^{2} + 24 ) ^{2}}

Compute the slope of

f (x) = \frac{125}{x ^{2} + 24}

x = - 1 : m = \frac{2}{5}

Find the line with slope m=

\frac{2}{5}

and passing through

(- 1, 5) : y = \frac{2}{5} x + \frac{27}{5}

y = \frac{2}{5} x + \frac{27}{5}

\frac{d}{d x} (cos (9 x^{2}))

\int \frac{1}{2 x ^{2} - 4 x + 20} d x

\frac{\partial}{\partial x} (x^{2} e^{y z^{2}})

x \to - 3.1 lim (x^{2})

\frac{d y}{d x} = \frac{x x ^{2} - y ^{2} + y ^{2}}{x y}

What is the tangent of f(x)=(125)/(x^2+24),\at x=-1 ?
The tangent of f(x)=(125)/(x^2+24),\at x=-1 is y= 2/5 x+27/5