What is the tangent of f(x)=4x-32sqrt(x),\at x=4 ?

The tangent of f(x)=4x-32sqrt(x),\at x=4 is y=-4x-32

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tangent of f(x)=4x-32sqrt(x),\at x=4

tangent of

f (x) = 4 x - 32 x, at x = 4

y = - 4 x - 32

Solution steps

Find the tangent point:

(4, - 48)

Find the slope of

f (x) = 4 x - 32 x : \frac{df ( x )}{d x} = 4 - \frac{16}{x}

Compute the slope of

f (x) = 4 x - 32 x

x = 4 : m = - 4

Find the line with slope m=

- 4

and passing through

(4, - 48) : y = - 4 x - 32

y = - 4 x - 32

x^{2} y^{^{'}} + 2 x y - y^{3} = 0

\int 1 + t^{2} d t

x \to - 1 lim (\frac{1}{x ^{2}})

- x^{3} (- 2.8)

\int \frac{x ^{2}}{e ^{2 x}} d x

What is the tangent of f(x)=4x-32sqrt(x),\at x=4 ?
The tangent of f(x)=4x-32sqrt(x),\at x=4 is y=-4x-32