What is the tangent of f(x)=4x^3-42x^2+146x+48 ?

The tangent of f(x)=4x^3-42x^2+146x+48 is y=(12a_{0}^2-84a_{0}+146)x-8a_{0}^3+42a_{0}^2+48

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tangent of f(x)=4x^3-42x^2+146x+48

tangent of

f (x) = 4 x^{3} - 42 x^{2} + 146 x + 48

y = (12 a_{0}^{2} - 84 a_{0} + 146) x - 8 a_{0}^{3} + 42 a_{0}^{2} + 48

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{3} - 42 a_{0}^{2} + 146 a_{0} + 48)

Find the slope of

f (x) = 4 x^{3} - 42 x^{2} + 146 x + 48 : \frac{df ( x )}{d x} = 12 x^{2} - 84 x + 146

Compute the slope of

f (x) = 4 x^{3} - 42 x^{2} + 146 x + 48

x = a_{0} : m = 12 a_{0}^{2} - 84 a_{0} + 146

Find the line with slope m=

12 a_{0}^{2} - 84 a_{0} + 146

and passing through

(a_{0}, 4 a_{0}^{3} - 42 a_{0}^{2} + 146 a_{0} + 48) : y = (12 a_{0}^{2} - 84 a_{0} + 146) x - 8 a_{0}^{3} + 42 a_{0}^{2} + 48

y = (12 a_{0}^{2} - 84 a_{0} + 146) x - 8 a_{0}^{3} + 42 a_{0}^{2} + 48

\frac{d}{d t} (4 sin (2 t))

\int \frac{3 x}{( x ^{2} + 5 ) ^{2}} d x

\frac{d}{d x} (\frac{11 - 11 x}{7 x ^{\frac{3}{7}}})

\frac{\partial}{\partial v} (\frac{1}{v})

\frac{d}{d x} (3 x^{2} - 12 x + 5)

What is the tangent of f(x)=4x^3-42x^2+146x+48 ?
The tangent of f(x)=4x^3-42x^2+146x+48 is y=(12a_{0}^2-84a_{0}+146)x-8a_{0}^3+42a_{0}^2+48