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

The tangent of f(x)= 1/(x^2),\at x=5 is y=-2/125 x+3/25

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

tangent of

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

y = - \frac{2}{125} x + \frac{3}{25}

Solution steps

Find the tangent point:

(5, \frac{1}{25})

Find the slope of

f (x) = \frac{1}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{2}{x ^{3}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{2}{125}

Find the line with slope m=

- \frac{2}{125}

and passing through

(5, \frac{1}{25}) : y = - \frac{2}{125} x + \frac{3}{25}

y = - \frac{2}{125} x + \frac{3}{25}

t an g e n t f (x) = 3 x^{2} - 4 x, (1, - 1)

t an g e n t - 3 cos (x)

\int cot^{4} (2 t) d t

\frac{d}{d x} (\frac{( x ^{2} - 4 ) ^{3}}{x ^{2} + 2})

f (X) = ln (ln (ln (tan (X))))

What is the tangent of f(x)= 1/(x^2),\at x=5 ?
The tangent of f(x)= 1/(x^2),\at x=5 is y=-2/125 x+3/25