What is the tangent of f(x)=sqrt(7-6x),\at x=1 ?

The tangent of f(x)=sqrt(7-6x),\at x=1 is y=-3x+4

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tangent of f(x)=sqrt(7-6x),\at x=1

tangent of

f (x) = 7 - 6 x, at x = 1

y = - 3 x + 4

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = 7 - 6 x : \frac{df ( x )}{d x} = - \frac{3}{7 - 6 x}

Compute the slope of

f (x) = 7 - 6 x

x = 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(1, 1) : y = - 3 x + 4

y = - 3 x + 4

y^{2} \frac{d y}{d x} = y^{3} - x^{3}

x \to - 2 lim (x^{2} + 5)

f (x) = x^{2}, g (x) = x + 2

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 9

f (x) = x^{2} + 2 x - 3

What is the tangent of f(x)=sqrt(7-6x),\at x=1 ?
The tangent of f(x)=sqrt(7-6x),\at x=1 is y=-3x+4