What is the tangent of f(x)=sqrt(2x^3+7x),\at x=1 ?

The tangent of f(x)=sqrt(2x^3+7x),\at x=1 is y= 13/6 x+5/6

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tangent of f(x)=sqrt(2x^3+7x),\at x=1

tangent of

f (x) = 2 x^{3} + 7 x, at x = 1

y = \frac{13}{6} x + \frac{5}{6}

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = 2 x^{3} + 7 x : \frac{df ( x )}{d x} = \frac{6 x ^{2} + 7}{2 2 x ^{3} + 7 x}

Compute the slope of

f (x) = 2 x^{3} + 7 x

x = 1 : m = \frac{13}{6}

Find the line with slope m=

\frac{13}{6}

and passing through

(1, 3) : y = \frac{13}{6} x + \frac{5}{6}

y = \frac{13}{6} x + \frac{5}{6}

\frac{d}{d x} (- 19 e^{x})

\int arcsin (2 x) d x

\int cos (4 x) (1 - sin (4 x))^{3} d x

x \to - 3 lim (\frac{x ^{2} - 8}{3 - x})

\frac{d}{d x} (e^{x^{2}} - x)

What is the tangent of f(x)=sqrt(2x^3+7x),\at x=1 ?
The tangent of f(x)=sqrt(2x^3+7x),\at x=1 is y= 13/6 x+5/6