What is the tangent of f(x)=sqrt(25-x),\at x=0 ?

The tangent of f(x)=sqrt(25-x),\at x=0 is y=-1/10 x+5

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tangent of f(x)=sqrt(25-x),\at x=0

tangent of

f (x) = 25 - x, at x = 0

y = - \frac{1}{10} x + 5

Solution steps

Find the tangent point:

(0, 5)

Find the slope of

f (x) = 25 - x : \frac{df ( x )}{d x} = - \frac{1}{2 25 - x}

Compute the slope of

f (x) = 25 - x

x = 0 : m = - \frac{1}{10}

Find the line with slope m=

- \frac{1}{10}

and passing through

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

y = - \frac{1}{10} x + 5

\int (3^{- x}) d x

\frac{d y}{d x}, y = x^{2} + 3

\frac{d}{d x} (x sin (\frac{π}{x}))

f (x) = tan (ln (x))

\int_{- 3}^{5} (2 y - (y^{2} - 15)) d y

What is the tangent of f(x)=sqrt(25-x),\at x=0 ?
The tangent of f(x)=sqrt(25-x),\at x=0 is y=-1/10 x+5