What is the tangent of f(x)=sqrt(2-10x),\at x=-1 ?

The tangent of f(x)=sqrt(2-10x),\at x=-1 is y=-5/(2sqrt(3))x+2sqrt(3)-5/(2sqrt(3))

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tangent of f(x)=sqrt(2-10x),\at x=-1

tangent of

f (x) = 2 - 10 x, at x = - 1

y = - \frac{5}{2 3} x + 23 - \frac{5}{2 3}

Solution steps

Find the tangent point:

(- 1, 23)

Find the slope of

f (x) = 2 - 10 x : \frac{df ( x )}{d x} = - \frac{5}{2 - 10 x}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{5}{2 3}

Find the line with slope m=

- \frac{5}{2 3}

and passing through

(- 1, 23) : y = - \frac{5}{2 3} x + 23 - \frac{5}{2 3}

y = - \frac{5}{2 3} x + 23 - \frac{5}{2 3}

\int (7 x) d x

\int \frac{1}{1 + u ^{2}} d u

\int \frac{x ^{3}}{( 1 + x ^{4} )} d x

x \to 0 lim (x^{2} + x - 3)

\int_{0}^{1} 2 π y (y - y^{2}) d y

What is the tangent of f(x)=sqrt(2-10x),\at x=-1 ?
The tangent of f(x)=sqrt(2-10x),\at x=-1 is y=-5/(2sqrt(3))x+2sqrt(3)-5/(2sqrt(3))