What is the tangent of f(x)=ln(x^3-7),\at x=2 ?

The tangent of f(x)=ln(x^3-7),\at x=2 is y=12x-24

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tangent of f(x)=ln(x^3-7),\at x=2

tangent of

f (x) = ln (x^{3} - 7), at x = 2

y = 12 x - 24

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

f (x) = ln (x^{3} - 7) : \frac{df ( x )}{d x} = \frac{3 x ^{2}}{x ^{3} - 7}

Compute the slope of

f (x) = ln (x^{3} - 7)

x = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 0) : y = 12 x - 24

y = 12 x - 24

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\int_{0}^{L} sin^{2} (\frac{nπ x}{L}) d x

y = \frac{e ^{x}}{1 + x}

\int_{0}^{\frac{π}{2}} 15 sin^{2} (2 x) d x

What is the tangent of f(x)=ln(x^3-7),\at x=2 ?
The tangent of f(x)=ln(x^3-7),\at x=2 is y=12x-24