What is the tangent of f(x)=x^3+7x-7,(1,1) ?

The tangent of f(x)=x^3+7x-7,(1,1) is y=10x-9

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tangent of f(x)=x^3+7x-7,(1,1)

tangent of

f (x) = x^{3} + 7 x - 7, (1, 1)

y = 10 x - 9

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 1) : m = 10

Find the line with slope m=

10

and passing through

(1, 1) : y = 10 x - 9

y = 10 x - 9

y^{^{''''}} = 0

y^{^{''}} - 6 y^{^{'}} + 25 y = 64 e^{- t}

\int 2 (x + 2)^{3} d x

\frac{\partial}{\partial x} (x^{3} + 3 x y^{2} - 15 x - 12 y)

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} - 24 y = 0

What is the tangent of f(x)=x^3+7x-7,(1,1) ?
The tangent of f(x)=x^3+7x-7,(1,1) is y=10x-9