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

The tangent of f(x)=8x^3-2x^2+2x,\at x=2 is y=90x-120

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tangent of f(x)=8x^3-2x^2+2x,\at x=2

tangent of

f (x) = 8 x^{3} - 2 x^{2} + 2 x, at x = 2

y = 90 x - 120

Solution steps

Find the tangent point:

(2, 60)

Find the slope of

f (x) = 8 x^{3} - 2 x^{2} + 2 x : \frac{df ( x )}{d x} = 24 x^{2} - 4 x + 2

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 90

Find the line with slope m=

90

and passing through

(2, 60) : y = 90 x - 120

y = 90 x - 120

\int_{0}^{t} sin (2 s) d s

\int \frac{2}{10 + 6 x + x ^{2}} d x

\frac{1}{4 x}

\int (1 - x) (1 - 2 x) (1 - 3 x) d x

\int \frac{sin ^{2} ( x ) cos ( 2 x )}{2} d x

What is the tangent of f(x)=8x^3-2x^2+2x,\at x=2 ?
The tangent of f(x)=8x^3-2x^2+2x,\at x=2 is y=90x-120