What is the tangent of f(x)= 5/(x^2),\at x=1 ?

The tangent of f(x)= 5/(x^2),\at x=1 is y=-10x+15

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tangent of f(x)= 5/(x^2),\at x=1

tangent of

f (x) = \frac{5}{x ^{2}}, at x = 1

y = - 10 x + 15

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

f (x) = \frac{5}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{10}{x ^{3}}

Compute the slope of

f (x) = \frac{5}{x ^{2}}

x = 1 : m = - 10

Find the line with slope m=

- 10

and passing through

(1, 5) : y = - 10 x + 15

y = - 10 x + 15

\int \frac{x - 1}{x ^{2} + 3 x + 2} d x

\frac{d y}{d x} = (3 y)^{\frac{1}{2}} e^{x + 10}

\int 10 x ln (x + 1) d x

\int_{- 1}^{1} (x^{3} - 9 x) d x

x \to 0 lim (\frac{1}{x} sin (\frac{x}{3}))

What is the tangent of f(x)= 5/(x^2),\at x=1 ?
The tangent of f(x)= 5/(x^2),\at x=1 is y=-10x+15