What is the tangent of f(x)=(x^2+8)(7-x) ?

The tangent of f(x)=(x^2+8)(7-x) is y=(14a_{0}-3a_{0}^2-8)x+2a_{0}^3-7a_{0}^2+56

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tangent of f(x)=(x^2+8)(7-x)

tangent of

f (x) = (x^{2} + 8) (7 - x)

y = (14 a_{0} - 3 a_{0}^{2} - 8) x + 2 a_{0}^{3} - 7 a_{0}^{2} + 56

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0}^{2} + 8) (7 - a_{0}))

Find the slope of

f (x) = (x^{2} + 8) (7 - x) : \frac{df ( x )}{d x} = 14 x - 3 x^{2} - 8

Compute the slope of

f (x) = (x^{2} + 8) (7 - x)

x = a_{0} : m = 14 a_{0} - 3 a_{0}^{2} - 8

Find the line with slope m=

14 a_{0} - 3 a_{0}^{2} - 8

and passing through

(a_{0}, (a_{0}^{2} + 8) (7 - a_{0})) : y = (14 a_{0} - 3 a_{0}^{2} - 8) x + 2 a_{0}^{3} - 7 a_{0}^{2} + 56

y = (14 a_{0} - 3 a_{0}^{2} - 8) x + 2 a_{0}^{3} - 7 a_{0}^{2} + 56

\int_{0}^{2} x^{7} e^{x^{4}} d x

\frac{d}{d x} ((2 x^{2} + 1)^{4})

x^{\frac{1}{4}}

f (x) = \frac{2 x + 2}{x + 7}

t cos (t)

What is the tangent of f(x)=(x^2+8)(7-x) ?
The tangent of f(x)=(x^2+8)(7-x) is y=(14a_{0}-3a_{0}^2-8)x+2a_{0}^3-7a_{0}^2+56