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

The tangent of f(x)=x^4-13x^2+36 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

tangent of

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

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_{0}^{1} \frac{45}{x ^{5}} d x

(1 + \frac{4}{x})^{x}

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

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

\int sin (20 x) sin (21 x) d x

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