What is the tangent of f(x)=x^4-17x^2+16,\at x=2 ?

The tangent of f(x)=x^4-17x^2+16,\at x=2 is y=-36x+36

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tangent of f(x)=x^4-17x^2+16,\at x=2

tangent of

f (x) = x^{4} - 17 x^{2} + 16, at x = 2

y = - 36 x + 36

Solution steps

Find the tangent point:

(2, - 36)

Find the slope of

f (x) = x^{4} - 17 x^{2} + 16 : \frac{df ( x )}{d x} = 4 x^{3} - 34 x

Compute the slope of

f (x) = x^{4} - 17 x^{2} + 16

x = 2 : m = - 36

Find the line with slope m=

- 36

and passing through

(2, - 36) : y = - 36 x + 36

y = - 36 x + 36

- x^{2} + 10, - 10, - 1

\int \frac{1}{sin ^{2} ( x ) \cdot cos ^{2} ( x )} d x

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3 x}}{x}

n = 1 \sum \infty 0

What is the tangent of f(x)=x^4-17x^2+16,\at x=2 ?
The tangent of f(x)=x^4-17x^2+16,\at x=2 is y=-36x+36