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

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

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

tangent of

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

y = 2 x - 1

Solution steps

Find the tangent point:

(2, 3)

Find the slope of

f (x) = 1 + x^{3} : \frac{df ( x )}{d x} = \frac{3 x ^{2}}{2 1 + x ^{3}}

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 = 2

Find the line with slope m=

2

and passing through

(2, 3) : y = 2 x - 1

y = 2 x - 1

\frac{d y}{d x} = x - 1

f (x) = \frac{6 x ^{2} + 10}{4 x ^{3}}

(e^{t})^{^{'}}

\int_{1}^{2} (x + 1)^{2} d x

e^{5 (t - 2)} (t^{3} + 2 t)^{2}

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