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

The tangent of f(x)=x^2,\at x=4 is y=8x-16

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

tangent of

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

y = 8 x - 16

Solution steps

Find the tangent point:

(4, 16)

Find the slope of

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

Compute the slope of

f (x) = x^{2}

x = 4 : m = 8

Find the line with slope m=

8

and passing through

(4, 16) : y = 8 x - 16

y = 8 x - 16

\frac{d}{d x} (2 + 3 e^{- \frac{3}{x}})

\frac{d}{d x} ((x^{4} + 3 x^{2} - 5)^{4})

\frac{d}{d x} (3 x e^{5 x})

\frac{d}{d x} (5^{- \frac{1}{x}})

3 x + 1

What is the tangent of f(x)=x^2,\at x=4 ?
The tangent of f(x)=x^2,\at x=4 is y=8x-16