What is the tangent of f(x)=5-x^2,\at x=1 ?

The tangent of f(x)=5-x^2,\at x=1 is y=-2x+6

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tangent of f(x)=5-x^2,\at x=1

tangent of

f (x) = 5 - x^{2}, at x = 1

y = - 2 x + 6

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = 5 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 5 - x^{2}

x = 1 : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 4) : y = - 2 x + 6

y = - 2 x + 6

x^{2} - 7

x \to 4 + lim (\frac{3}{x - 4})

2 y^{^{'}} - y^{2} + 4 = 0

\int \frac{x ^{3}}{1 - 2 x ^{2}} d x

3 x e^{3 x}

What is the tangent of f(x)=5-x^2,\at x=1 ?
The tangent of f(x)=5-x^2,\at x=1 is y=-2x+6