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

The tangent of f(x)=(1+2x)^2,\at x=3 is y=28x-35

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tangent of f(x)=(1+2x)^2,\at x=3

tangent of

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

y = 28 x - 35

Solution steps

Find the tangent point:

(3, 49)

Find the slope of

f (x) = (1 + 2 x)^{2} : \frac{df ( x )}{d x} = 4 (1 + 2 x)

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 28

Find the line with slope m=

28

and passing through

(3, 49) : y = 28 x - 35

y = 28 x - 35

g

t^{2} + 5

\frac{d}{d x} (\frac{1}{( x ^{5} - x + 1 ) ^{9}})

y^{^{''}} - 26 y^{^{'}} + 169 y = 0

\int 10 x + 11 x d x

What is the tangent of f(x)=(1+2x)^2,\at x=3 ?
The tangent of f(x)=(1+2x)^2,\at x=3 is y=28x-35