What is the tangent of f(x)=sqrt(32-x) ?

The tangent of f(x)=sqrt(32-x) is y=-1/(2sqrt(32-a_{0))}x+sqrt(32-a_{0)}+(a_{0})/(2sqrt(32-a_{0))}

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tangent of f(x)=sqrt(32-x)

tangent of

f (x) = 32 - x

y = - \frac{1}{2 32 - a _{0}} x + 32 - a_{0} + \frac{a _{0}}{2 32 - a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 32 - a_{0})

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}{2 32 - a _{0}}

Find the line with slope m=

- \frac{1}{2 32 - a _{0}}

and passing through

(a_{0}, 32 - a_{0}) : y = - \frac{1}{2 32 - a _{0}} x + 32 - a_{0} + \frac{a _{0}}{2 32 - a _{0}}

y = - \frac{1}{2 32 - a _{0}} x + 32 - a_{0} + \frac{a _{0}}{2 32 - a _{0}}

y^{^{'}} - 2 \cdot y = e^{(2 \cdot t)}

\frac{d}{d x} (\frac{3 x + 2}{4 x})

\frac{d}{d x} (- sin (7 x^{2}) \cdot 14 x)

\frac{d x}{d t} + 2 t x = t, x (0) = \frac{3}{2}

f (x) = 5 x^{3} + 4 x, at x = - 3

What is the tangent of f(x)=sqrt(32-x) ?
The tangent of f(x)=sqrt(32-x) is y=-1/(2sqrt(32-a_{0))}x+sqrt(32-a_{0)}+(a_{0})/(2sqrt(32-a_{0))}