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

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

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

tangent of

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

y = 3 x - 1

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

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 = 3

Find the line with slope m=

3

and passing through

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

y = 3 x - 1

\frac{d}{d o} (ln (o))

x \to - 6 lim (\frac{- 5 x}{x + 6})

d er i v a t i v e \frac{1}{2 x} + \frac{1}{6 x ^{\frac{5}{6}}}

\frac{d}{d x} (sin (x + sin (10 x)))

\int (x - x^{3}) d x

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