What is the tangent of f(x)=2x^3-5x+4,\at x=-1 ?

The tangent of f(x)=2x^3-5x+4,\at x=-1 is y=x+8

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tangent of f(x)=2x^3-5x+4,\at x=-1

tangent of

f (x) = 2 x^{3} - 5 x + 4, at x = - 1

y = x + 8

Solution steps

Find the tangent point:

(- 1, 7)

Find the slope of

f (x) = 2 x^{3} - 5 x + 4 : \frac{df ( x )}{d x} = 6 x^{2} - 5

Compute the slope of

f (x) = 2 x^{3} - 5 x + 4

x = - 1 : m = 1

Find the line with slope m=

1

and passing through

(- 1, 7) : y = x + 8

y = x + 8

\frac{1}{3 x ^{3}} + \frac{x ^{7}}{10}

x \to 2 + lim (x - 1)

\frac{5 s}{( s + 4 ) ^{3}}

\frac{\partial}{\partial x} (\frac{2 x}{( x ^{2} + y ^{2} )})

(2 x + 5) (5 x^{2} + 1)

What is the tangent of f(x)=2x^3-5x+4,\at x=-1 ?
The tangent of f(x)=2x^3-5x+4,\at x=-1 is y=x+8