What is the tangent of f(x)=x^4+5x^2-x,\at x=1 ?

The tangent of f(x)=x^4+5x^2-x,\at x=1 is y=13x-8

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tangent of f(x)=x^4+5x^2-x,\at x=1

tangent of

f (x) = x^{4} + 5 x^{2} - x, at x = 1

y = 13 x - 8

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

f (x) = x^{4} + 5 x^{2} - x : \frac{df ( x )}{d x} = 4 x^{3} + 10 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 = 13

Find the line with slope m=

13

and passing through

(1, 5) : y = 13 x - 8

y = 13 x - 8

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

\int ln (cos (x)) \cdot tan (x) d x

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

n = 0 \sum \infty (- \frac{8}{9})^{n}

x^{^{'}} = k x (t)

What is the tangent of f(x)=x^4+5x^2-x,\at x=1 ?
The tangent of f(x)=x^4+5x^2-x,\at x=1 is y=13x-8