What is the tangent of f(x)=6x^2+4/(x^2)-10,\at x=1 ?

The tangent of f(x)=6x^2+4/(x^2)-10,\at x=1 is y=4x-4

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tangent of f(x)=6x^2+4/(x^2)-10,\at x=1

tangent of

f (x) = 6 x^{2} + \frac{4}{x ^{2}} - 10, at x = 1

y = 4 x - 4

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = 6 x^{2} + \frac{4}{x ^{2}} - 10 : \frac{df ( x )}{d x} = 12 x - \frac{8}{x ^{3}}

Compute the slope of

f (x) = 6 x^{2} + \frac{4}{x ^{2}} - 10

x = 1 : m = 4

Find the line with slope m=

4

and passing through

(1, 0) : y = 4 x - 4

y = 4 x - 4

\int 5 sin (5 x) d x

\int \frac{3 x ^{2} + x + 4}{x ^{4} + 3 x ^{2} + 2} d x

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

\frac{d}{d x} (x^{4} ln (9 x))

What is the tangent of f(x)=6x^2+4/(x^2)-10,\at x=1 ?
The tangent of f(x)=6x^2+4/(x^2)-10,\at x=1 is y=4x-4