What is the tangent of f(x)=x^3-9x^2+10x,\at x=1 ?

The tangent of f(x)=x^3-9x^2+10x,\at x=1 is y=-5x+7

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tangent of f(x)=x^3-9x^2+10x,\at x=1

tangent of

f (x) = x^{3} - 9 x^{2} + 10 x, at x = 1

y = - 5 x + 7

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = x^{3} - 9 x^{2} + 10 x : \frac{df ( x )}{d x} = 3 x^{2} - 18 x + 10

Compute the slope of

f (x) = x^{3} - 9 x^{2} + 10 x

x = 1 : m = - 5

Find the line with slope m=

- 5

and passing through

(1, 2) : y = - 5 x + 7

y = - 5 x + 7

x^{2} - 5 x, at x = 4

\frac{d}{d x} (arctan (x^{3}))

e^{- s t} d t

\int 7 sin^{3} (x co) s^{7} x d x

\int (z + 1) e^{6 z} d z

What is the tangent of f(x)=x^3-9x^2+10x,\at x=1 ?
The tangent of f(x)=x^3-9x^2+10x,\at x=1 is y=-5x+7