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

The tangent of f(x)=x^4+2x^2-x,\at x=1 is y=7x-5

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

tangent of

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

y = 7 x - 5

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 7

Find the line with slope m=

7

and passing through

(1, 2) : y = 7 x - 5

y = 7 x - 5

x^{2} y^{^{'}} + x y = 5

\frac{d}{d x} (9 x^{4} + 6 x^{3} + 1)

\int ln (x + 1 + x) d x

lo g_{10} (x^{7} + 5)

\frac{d}{d x} (\frac{1}{3 x ^{\frac{2}{3}}})

What is the tangent of f(x)=x^4+2x^2-x,\at x=1 ?
The tangent of f(x)=x^4+2x^2-x,\at x=1 is y=7x-5