What is the tangent of f(x)=x^3-3x-5,\at x=1 ?

The tangent of f(x)=x^3-3x-5,\at x=1 is y=-7

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tangent of f(x)=x^3-3x-5,\at x=1

tangent of

f (x) = x^{3} - 3 x - 5, at x = 1

y = - 7

Solution steps

Find the tangent point:

(1, - 7)

Find the slope of

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

Compute the slope of

f (x) = x^{3} - 3 x - 5

x = 1 : m = 0

Find the line with slope m=

0

and passing through

(1, - 7) : y = - 7

y = - 7

\frac{\partial}{\partial x} (ln (x + 6 y + 8 z))

\frac{d y}{d x} = - \frac{2}{x ^{3}} + 2 e^{2 x - 2}

f (x) = 2 x^{2} + 5 x^{2} - 2, at x = 2

y = x^{2} + 6

\int_{- 1}^{1} x^{3} - x d x

What is the tangent of f(x)=x^3-3x-5,\at x=1 ?
The tangent of f(x)=x^3-3x-5,\at x=1 is y=-7