What is the tangent of f(x)=4-2x^2,\at x=5 ?

The tangent of f(x)=4-2x^2,\at x=5 is y=-20x+54

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tangent of f(x)=4-2x^2,\at x=5

tangent of

f (x) = 4 - 2 x^{2}, at x = 5

y = - 20 x + 54

Solution steps

Find the tangent point:

(5, - 46)

Find the slope of

f (x) = 4 - 2 x^{2} : \frac{df ( x )}{d x} = - 4 x

Compute the slope of

f (x) = 4 - 2 x^{2}

x = 5 : m = - 20

Find the line with slope m=

- 20

and passing through

(5, - 46) : y = - 20 x + 54

y = - 20 x + 54

(\frac{- 2 x - 1}{- 4 x + 2})^{5}

\frac{\partial}{\partial y} (x^{2} + x y^{4})

\int_{1}^{x} \frac{1}{t} d t

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

y = \frac{x - x}{x ^{2}}

What is the tangent of f(x)=4-2x^2,\at x=5 ?
The tangent of f(x)=4-2x^2,\at x=5 is y=-20x+54