What is the tangent of f(x)=8xtan(x),\at x=pi ?

The tangent of f(x)=8xtan(x),\at x=pi is y=8pix-8pi^2

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tangent of f(x)=8xtan(x),\at x=pi

tangent of

f (x) = 8 x tan (x), at x = π

y = 8 π x - 8 π^{2}

Solution steps

Find the tangent point:

(π, 0)

Find the slope of

f (x) = 8 x tan (x) : \frac{df ( x )}{d x} = 8 (tan (x) + x sec^{2} (x))

Compute the slope of

f (x) = 8 x tan (x)

x = π : m = 8 π

Find the line with slope m=

8 π

and passing through

(π, 0) : y = 8 π x - 8 π^{2}

y = 8 π x - 8 π^{2}

f (x) = 3 - x

4 y^{^{''}} - 20 y^{^{'}} + 25 y = 0

\frac{d}{d x} (x^{2} 4 - x)

(- 3.1) (2.1)

x^{^{'}} = 6 - \frac{3}{5} \cdot x

What is the tangent of f(x)=8xtan(x),\at x=pi ?
The tangent of f(x)=8xtan(x),\at x=pi is y=8pix-8pi^2