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

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

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

tangent of

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

y = 8 x - 5

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 8

Find the line with slope m=

8

and passing through

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

y = 8 x - 5

\frac{d y}{d x} (x^{2} + 3) = x y

\int \frac{16 x ^{2} - 49}{x ^{3}} d x

\frac{( y - 3 )}{y ^{2} - 3 y + 9}

y^{^{''}} + 4 y = t - 2 sin (2 t)

\frac{d}{d y} (x^{2} y sin^{3} (x y^{2}))

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