What is the tangent of f(x)= 3/(x^2),\at x=1 ?

The tangent of f(x)= 3/(x^2),\at x=1 is y=-6x+9

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tangent of f(x)= 3/(x^2),\at x=1

tangent of

f (x) = \frac{3}{x ^{2}}, at x = 1

y = - 6 x + 9

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = \frac{3}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{6}{x ^{3}}

Compute the slope of

f (x) = \frac{3}{x ^{2}}

x = 1 : m = - 6

Find the line with slope m=

- 6

and passing through

(1, 3) : y = - 6 x + 9

y = - 6 x + 9

\int_{0}^{\frac{π}{4}} 4 sin^{2} (x) d x

h \to 0 lim ((1 + 5 h)^{\frac{1}{h}})

\frac{d}{d x} (y^{2} - x^{2})

\int e^{5 x + e^{5 x}} d x

\frac{s}{2 s ^{2} + 4 s + 3}

What is the tangent of f(x)= 3/(x^2),\at x=1 ?
The tangent of f(x)= 3/(x^2),\at x=1 is y=-6x+9