What is the tangent of f(x)=x^4-18x^2+81,\at x=1 ?

The tangent of f(x)=x^4-18x^2+81,\at x=1 is y=-32x+96

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tangent of f(x)=x^4-18x^2+81,\at x=1

tangent of

f (x) = x^{4} - 18 x^{2} + 81, at x = 1

y = - 32 x + 96

Solution steps

Find the tangent point:

(1, 64)

Find the slope of

f (x) = x^{4} - 18 x^{2} + 81 : \frac{df ( x )}{d x} = 4 x^{3} - 36 x

Compute the slope of

f (x) = x^{4} - 18 x^{2} + 81

x = 1 : m = - 32

Find the line with slope m=

- 32

and passing through

(1, 64) : y = - 32 x + 96

y = - 32 x + 96

\frac{\partial}{\partial x} (2 y^{3} x)

\frac{d}{d x} (sin (\frac{x}{y}))

\frac{d}{d x} (\frac{17}{x})

\int \frac{sec ^{2} ( x )}{sec ( x )} d x

\frac{d}{d x} (e^{y x})

What is the tangent of f(x)=x^4-18x^2+81,\at x=1 ?
The tangent of f(x)=x^4-18x^2+81,\at x=1 is y=-32x+96