What is the tangent of f(x)=sqrt(7x+32),\at x=-1 ?

The tangent of f(x)=sqrt(7x+32),\at x=-1 is y= 7/10 x+57/10

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tangent of f(x)=sqrt(7x+32),\at x=-1

tangent of

f (x) = 7 x + 32, at x = - 1

y = \frac{7}{10} x + \frac{57}{10}

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

f (x) = 7 x + 32 : \frac{df ( x )}{d x} = \frac{7}{2 7 x + 32}

Compute the slope of

f (x) = 7 x + 32

x = - 1 : m = \frac{7}{10}

Find the line with slope m=

\frac{7}{10}

and passing through

(- 1, 5) : y = \frac{7}{10} x + \frac{57}{10}

y = \frac{7}{10} x + \frac{57}{10}

(x + y) d x + (x + y - 1) d y = 0

\int (\frac{- y}{x ^{2} + y ^{2}}) d x

\frac{\partial}{\partial x} ((x - y) sin (3 x + 2 y))

\frac{6}{( s ^{2} ( s ^{2} + 8 s + 1 ) )}

y = ln ∣ x ∣

What is the tangent of f(x)=sqrt(7x+32),\at x=-1 ?
The tangent of f(x)=sqrt(7x+32),\at x=-1 is y= 7/10 x+57/10