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

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

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

tangent of

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

y = - \frac{1}{10} x + \frac{57}{10}

Solution steps

Find the tangent point:

(7, 5)

Find the slope of

f (x) = 32 - x : \frac{df ( x )}{d x} = - \frac{1}{2 32 - x}

Compute the slope of

f (x) = 32 - x

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

Find the line with slope m=

- \frac{1}{10}

and passing through

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

y = - \frac{1}{10} x + \frac{57}{10}

4 x^{3}

\frac{d}{d t} (\frac{3}{t})

f (x) = 4 tan (x)

\frac{d}{d x} (2 arctan (sin (x)))

x^{2} + y^{2} = 25, (3, - 4)

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