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

The tangent of f(x)=3x^3-x^2+4,\at x=1 is y=7x-1

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

tangent of

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

y = 7 x - 1

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = 3 x^{3} - x^{2} + 4 : \frac{df ( x )}{d x} = 9 x^{2} - 2 x

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 = 7

Find the line with slope m=

7

and passing through

(1, 6) : y = 7 x - 1

y = 7 x - 1

\frac{d}{d x} (e^{4 x^{2}} csc (5 x))

sin (- 2 x)

\int \frac{1}{10 ( x ^{2} + 1 )} d x

(2 x - 7)^{- 1} (x + 5)

\frac{d}{d x} (- 4 cot (x))

What is the tangent of f(x)=3x^3-x^2+4,\at x=1 ?
The tangent of f(x)=3x^3-x^2+4,\at x=1 is y=7x-1