What is the tangent of f(x)=x^4-13x^2+36,\at 1 ?

The tangent of f(x)=x^4-13x^2+36,\at 1 is y=(4a_{0}^3-26a_{0})x-3a_{0}^4+13a_{0}^2+36

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tangent of f(x)=x^4-13x^2+36,\at 1

tangent of

f (x) = x^{4} - 13 x^{2} + 36, at 1

y = (4 a_{0}^{3} - 26 a_{0}) x - 3 a_{0}^{4} + 13 a_{0}^{2} + 36

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 13 a_{0}^{2} + 36)

Find the slope of

f (x) = x^{4} - 13 x^{2} + 36 : \frac{df ( x )}{d x} = 4 x^{3} - 26 x

Compute the slope of

f (x) = x^{4} - 13 x^{2} + 36

x = a_{0} : m = 4 a_{0}^{3} - 26 a_{0}

Find the line with slope m=

4 a_{0}^{3} - 26 a_{0}

and passing through

(a_{0}, a_{0}^{4} - 13 a_{0}^{2} + 36) : y = (4 a_{0}^{3} - 26 a_{0}) x - 3 a_{0}^{4} + 13 a_{0}^{2} + 36

y = (4 a_{0}^{3} - 26 a_{0}) x - 3 a_{0}^{4} + 13 a_{0}^{2} + 36

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

\int \frac{5 y ^{4}}{y ^{5} - 13} d y

\frac{50 s}{s ^{2} + 25}

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

\int_{- 2}^{5} - x^{3} + 9 x d x

What is the tangent of f(x)=x^4-13x^2+36,\at 1 ?
The tangent of f(x)=x^4-13x^2+36,\at 1 is y=(4a_{0}^3-26a_{0})x-3a_{0}^4+13a_{0}^2+36