What is the tangent of f(x)=10x^3-45x^2+60x-30 ?

The tangent of f(x)=10x^3-45x^2+60x-30 is y=(30a_{0}^2-90a_{0}+60)x-20a_{0}^3+45a_{0}^2-30

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tangent of f(x)=10x^3-45x^2+60x-30

tangent of

f (x) = 10 x^{3} - 45 x^{2} + 60 x - 30

y = (30 a_{0}^{2} - 90 a_{0} + 60) x - 20 a_{0}^{3} + 45 a_{0}^{2} - 30

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 10 a_{0}^{3} - 45 a_{0}^{2} + 60 a_{0} - 30)

Find the slope of

f (x) = 10 x^{3} - 45 x^{2} + 60 x - 30 : \frac{df ( x )}{d x} = 30 x^{2} - 90 x + 60

Compute the slope of

f (x) = 10 x^{3} - 45 x^{2} + 60 x - 30

x = a_{0} : m = 30 a_{0}^{2} - 90 a_{0} + 60

Find the line with slope m=

30 a_{0}^{2} - 90 a_{0} + 60

and passing through

(a_{0}, 10 a_{0}^{3} - 45 a_{0}^{2} + 60 a_{0} - 30) : y = (30 a_{0}^{2} - 90 a_{0} + 60) x - 20 a_{0}^{3} + 45 a_{0}^{2} - 30

y = (30 a_{0}^{2} - 90 a_{0} + 60) x - 20 a_{0}^{3} + 45 a_{0}^{2} - 30

\frac{d}{d s} (\frac{( s ^{2} - 4 s + 20 )}{s ^{2} + 6 s + 8})

y = e^{2 x}, x = 0, x = 3

\int \frac{e ^{6 t}}{t} d t

\frac{d}{d x} (e^{(\frac{- k}{x})})

2 (sin (x^{2}) + 2 x^{2} cos (x^{2}))

What is the tangent of f(x)=10x^3-45x^2+60x-30 ?
The tangent of f(x)=10x^3-45x^2+60x-30 is y=(30a_{0}^2-90a_{0}+60)x-20a_{0}^3+45a_{0}^2-30