What is the tangent of f(x)=-3x^4+7x^2-4,\at x=1 ?

The tangent of f(x)=-3x^4+7x^2-4,\at x=1 is y=2x-2

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tangent of f(x)=-3x^4+7x^2-4,\at x=1

tangent of

f (x) = - 3 x^{4} + 7 x^{2} - 4, at x = 1

y = 2 x - 2

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = - 3 x^{4} + 7 x^{2} - 4 : \frac{df ( x )}{d x} = - 12 x^{3} + 14 x

Compute the slope of

f (x) = - 3 x^{4} + 7 x^{2} - 4

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 0) : y = 2 x - 2

y = 2 x - 2

\frac{\partial}{\partial x} (2 x - 5 y)

\frac{1}{s ^{2} + 0.1 s + 1}

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

x \to 0 lim (x (x - [x]))

\int x cos (\frac{nπ x}{5}) d x

What is the tangent of f(x)=-3x^4+7x^2-4,\at x=1 ?
The tangent of f(x)=-3x^4+7x^2-4,\at x=1 is y=2x-2