What is the tangent of f(x)=x^{[2]}+1,\at x=3 ?

The tangent of f(x)=x^{[2]}+1,\at x=3 is y=6x-8

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tangent of f(x)=x^{[2]}+1,\at x=3

tangent of

f (x) = x^{[2]} + 1, at x = 3

y = 6 x - 8

Solution steps

Find the tangent point:

(3, 10)

Find the slope of

f (x) = x^{[2]} + 1 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{[2]} + 1

x = 3 : m = 6

Find the line with slope m=

6

and passing through

(3, 10) : y = 6 x - 8

y = 6 x - 8

\int \frac{3}{cos ^{2} ( 3 x )} d x

n = 1 \sum \infty (\frac{1}{11})^{n}

\int x \cdot sin (a x) d x

n = 0 \sum \infty \frac{2 ^{n}}{100}

x^{2} y^{^{''}} + x y^{^{'}} - y = \frac{1}{x + 1}

What is the tangent of f(x)=x^{[2]}+1,\at x=3 ?
The tangent of f(x)=x^{[2]}+1,\at x=3 is y=6x-8