What is the tangent of f(x)=4-x^2,\at x=-1 ?

The tangent of f(x)=4-x^2,\at x=-1 is y=2x+5

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tangent of f(x)=4-x^2,\at x=-1

tangent of

f (x) = 4 - x^{2}, at x = - 1

y = 2 x + 5

Solution steps

Find the tangent point:

(- 1, 3)

Find the slope of

f (x) = 4 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 4 - x^{2}

x = - 1 : m = 2

Find the line with slope m=

2

and passing through

(- 1, 3) : y = 2 x + 5

y = 2 x + 5

\int \frac{1}{10 x ^{7}} d x

y^{'} = y (3 - y)

\int (cos (y) sin (x)) d y

\int \frac{9 sec ^{2} ( x )}{tan ( x )} d x

x \to 2 lim (x + x)

What is the tangent of f(x)=4-x^2,\at x=-1 ?
The tangent of f(x)=4-x^2,\at x=-1 is y=2x+5