What is the tangent of f(x)=x^2,\at x=3,x=7 ?

The tangent of f(x)=x^2,\at x=3,x=7 is y=6x-9

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tangent of f(x)=x^2,\at x=3,x=7

tangent of

f (x) = x^{2}, at x = 3, x = 7

y = 6 x - 9

Solution steps

Find the tangent point:

(3, 9)

Find the slope of

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

Compute the slope of

f (x) = x^{2}

x = 3 : m = 6

Find the line with slope m=

6

and passing through

(3, 9) : y = 6 x - 9

y = 6 x - 9

\int_{1}^{5} \frac{4}{x ^{2} + 4 x + 4} d x

\frac{d v}{d t} = \frac{1}{t ^{2}}

x \to 8 - lim (\frac{x ^{8}}{x - 8})

\frac{d}{d x} (30 e^{- \frac{x ^{2}}{3}})

\frac{\partial}{\partial y} (e^{y} (y^{2} - x^{2}))

What is the tangent of f(x)=x^2,\at x=3,x=7 ?
The tangent of f(x)=x^2,\at x=3,x=7 is y=6x-9