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

The tangent of f(x)=x^4+3x^2-x,(1,3) is y=9x-6

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

tangent of

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

y = 9 x - 6

Solution steps

Find the slope of

f (x) = x^{4} + 3 x^{2} - x : \frac{df ( x )}{d x} = 4 x^{3} + 6 x - 1

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 = 9

Find the line with slope m=

9

and passing through

(1, 3) : y = 9 x - 6

y = 9 x - 6

\int_{0}^{a} x d x

\int_{0}^{1} 6 x^{2} + 4 x - 3 + 4 e^{4 x} d x

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{t}

(5, 2), (2, - 3)

\frac{d}{d x} (\frac{x}{x ^{4} + 1})

What is the tangent of f(x)=x^4+3x^2-x,(1,3) ?
The tangent of f(x)=x^4+3x^2-x,(1,3) is y=9x-6