What is the tangent of f(x)=x^4-20x^2+64,\at x=1 ?

The tangent of f(x)=x^4-20x^2+64,\at x=1 is y=-36x+81

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tangent of f(x)=x^4-20x^2+64,\at x=1

tangent of

f (x) = x^{4} - 20 x^{2} + 64, at x = 1

y = - 36 x + 81

Solution steps

Find the tangent point:

(1, 45)

Find the slope of

f (x) = x^{4} - 20 x^{2} + 64 : \frac{df ( x )}{d x} = 4 x^{3} - 40 x

Compute the slope of

f (x) = x^{4} - 20 x^{2} + 64

x = 1 : m = - 36

Find the line with slope m=

- 36

and passing through

(1, 45) : y = - 36 x + 81

y = - 36 x + 81

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

\int 2 x - 3 d x

f (x) = - x + \frac{7}{x} + 1 at x

(x^{2} \cdot 5^{x})^{7}

x \to 2 lim (\frac{x ^{2} + 4}{x + 2})

What is the tangent of f(x)=x^4-20x^2+64,\at x=1 ?
The tangent of f(x)=x^4-20x^2+64,\at x=1 is y=-36x+81