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}

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{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}

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