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

The tangent of y=f(x),\at x=2 is y= d/(d2)(f(2))x+f(2)-d/(d2)(f(2))2

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tangent of y=f(x),\at x=2

tangent of

y = f (x), at x = 2

y = \frac{d}{d 2} (f (2)) x + f (2) - \frac{d}{d 2} (f (2)) 2

Solution steps

Find the tangent point:

(2, f (2))

Find the slope of

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

Compute the slope of

y = f (x)

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

Find the line with slope m=

\frac{d}{d 2} (f (2))

and passing through

(2, f (2)) : y = \frac{d}{d 2} (f (2)) x + f (2) - \frac{d}{d 2} (f (2)) 2

y = \frac{d}{d 2} (f (2)) x + f (2) - \frac{d}{d 2} (f (2)) 2

x \to - 2 - lim (\frac{- 3}{x ^{2} - 4})

n = 1 \sum \infty \frac{1}{n ^{3} + 2}

f (x) = x^{2} (1 - 7 x)

\int \frac{2}{x ( x ^{2} + 25 )} d x

cos^{2} (x) sin (x) y^{^{'}} + cos^{3} (x) y = 2

What is the tangent of y=f(x),\at x=2 ?
The tangent of y=f(x),\at x=2 is y= d/(d2)(f(2))x+f(2)-d/(d2)(f(2))2