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

The tangent of f(x)=x^4+7x^2-x,\at x=1 is y=17x-10

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

tangent of

f (x) = x^{4} + 7 x^{2} - x, at x = 1

y = 17 x - 10

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

f (x) = x^{4} + 7 x^{2} - x : \frac{df ( x )}{d x} = 4 x^{3} + 14 x - 1

Compute the slope of

f (x) = x^{4} + 7 x^{2} - x

x = 1 : m = 17

Find the line with slope m=

17

and passing through

(1, 7) : y = 17 x - 10

y = 17 x - 10

\frac{( x ^{2} + 1 )}{( x + 1 )}, 1

\frac{\partial}{\partial u} (u - uv)

\frac{2}{x ^{4}}

f (y) = y ln (1 + e^{y})

\int \frac{cos ( x )}{sin ^{3} ( x ) - cos ^{3} ( x )} d x

What is the tangent of f(x)=x^4+7x^2-x,\at x=1 ?
The tangent of f(x)=x^4+7x^2-x,\at x=1 is y=17x-10