What is the tangent of f(x)=1+4csc(x),\at x=-pi/3 ?

The tangent of f(x)=1+4csc(x),\at x=-pi/3 is y=-8/3 x+1-8/(sqrt(3))-(8pi)/9

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tangent of f(x)=1+4csc(x),\at x=-pi/3

tangent of

f (x) = 1 + 4 csc (x), at x = - \frac{π}{3}

y = - \frac{8}{3} x + 1 - \frac{8}{3} - \frac{8 π}{9}

Solution steps

Find the tangent point:

(- \frac{π}{3}, 1 - \frac{8}{3})

Find the slope of

f (x) = 1 + 4 csc (x) : \frac{df ( x )}{d x} = - 4 cot (x) csc (x)

Compute the slope of

f (x) = 1 + 4 csc (x)

x = - \frac{π}{3} : m = - \frac{8}{3}

Find the line with slope m=

- \frac{8}{3}

and passing through

(- \frac{π}{3}, 1 - \frac{8}{3}) : y = - \frac{8}{3} x + 1 - \frac{8}{3} - \frac{8 π}{9}

y = - \frac{8}{3} x + 1 - \frac{8}{3} - \frac{8 π}{9}

n = 1 \sum \infty \frac{2}{3 n + 10}

\int_{1}^{2} 4 x 4 x^{2} + 4 d x

\int (3 x^{4} - 5 x^{2} + 6 x) d x

x \to 0 lim (\frac{∣ x + 1 ∣ - ∣ x - 1 ∣}{x})

n = 1 \sum \infty 8

What is the tangent of f(x)=1+4csc(x),\at x=-pi/3 ?
The tangent of f(x)=1+4csc(x),\at x=-pi/3 is y=-8/3 x+1-8/(sqrt(3))-(8pi)/9