What is the tangent of f(x)=sqrt(-x^2+25),\at x=3 ?

The tangent of f(x)=sqrt(-x^2+25),\at x=3 is y=-3/4 x+25/4

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tangent of f(x)=sqrt(-x^2+25),\at x=3

tangent of

f (x) = - x^{2} + 25, at x = 3

y = - \frac{3}{4} x + \frac{25}{4}

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

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

Compute the slope of

f (x) = - x^{2} + 25

x = 3 : m = - \frac{3}{4}

Find the line with slope m=

- \frac{3}{4}

and passing through

(3, 4) : y = - \frac{3}{4} x + \frac{25}{4}

y = - \frac{3}{4} x + \frac{25}{4}

\int (1 + 2 x)^{- 1} d x

\int_{0}^{\infty} \frac{1}{x ^{2} + 4} d x

x^{4} e^{- x}

y = e^{2 e x}

\frac{\partial}{\partial x} (x^{3} - y^{3} + 6 x y)

What is the tangent of f(x)=sqrt(-x^2+25),\at x=3 ?
The tangent of f(x)=sqrt(-x^2+25),\at x=3 is y=-3/4 x+25/4