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

The tangent of f(x)=x^4-2x^2+1,\at x=2 is y=24x-39

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

tangent of

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

y = 24 x - 39

Solution steps

Find the tangent point:

(2, 9)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 24

Find the line with slope m=

24

and passing through

(2, 9) : y = 24 x - 39

y = 24 x - 39

\int_{0}^{π} sin (x) sin (3 x) d x

\frac{x ^{2}}{x ^{2} + 3}

y^{^{'}} = x^{2} y - y + x^{2} - 1

\int e [2 x] d x

\int \frac{3 x}{x ^{2} - 1} d x

What is the tangent of f(x)=x^4-2x^2+1,\at x=2 ?
The tangent of f(x)=x^4-2x^2+1,\at x=2 is y=24x-39