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

The tangent of f(x)=x^4+4x^2-x,\at x=1 is y=11x-7

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

tangent of

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

y = 11 x - 7

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = x^{4} + 4 x^{2} - x : \frac{df ( x )}{d x} = 4 x^{3} + 8 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 = 11

Find the line with slope m=

11

and passing through

(1, 4) : y = 11 x - 7

y = 11 x - 7

\int sin^{5} (7 x) cos^{2} (7 x) d x

\frac{d}{d x} (2 x - 3 3 x)

\frac{d}{d x} (a sin (x) + b cos (x))

\int \frac{ln ( x ^{7} )}{x} d x

\int \frac{1}{( 4 - x ) ^{2}} d x

What is the tangent of f(x)=x^4+4x^2-x,\at x=1 ?
The tangent of f(x)=x^4+4x^2-x,\at x=1 is y=11x-7