What is the tangent of y= 4/3 x^3-5x+4 ?

The tangent of y= 4/3 x^3-5x+4 is y=(4a_{0}^2-5)x-8/3 a_{0}^3+4

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tangent of y= 4/3 x^3-5x+4

tangent of

y = \frac{4}{3} x^{3} - 5 x + 4

y = (4 a_{0}^{2} - 5) x - \frac{8}{3} a_{0}^{3} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{4}{3} a_{0}^{3} - 5 a_{0} + 4)

Find the slope of

y = \frac{4}{3} x^{3} - 5 x + 4 : \frac{d y}{d x} = 4 x^{2} - 5

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 = 4 a_{0}^{2} - 5

Find the line with slope m=

4 a_{0}^{2} - 5

and passing through

(a_{0}, \frac{4}{3} a_{0}^{3} - 5 a_{0} + 4) : y = (4 a_{0}^{2} - 5) x - \frac{8}{3} a_{0}^{3} + 4

y = (4 a_{0}^{2} - 5) x - \frac{8}{3} a_{0}^{3} + 4

\frac{d}{d x} (4 x 3 x^{4})

\int \frac{x ^{2}}{( 9 - 25 x ^{2} ) ^{\frac{5}{2}}} d x

\frac{1 - x}{1 + x}

\frac{d}{d x} (\frac{1}{5} e^{x})

x \to 2 lim (\frac{2 ^{x} - 4}{x - 2})

What is the tangent of y= 4/3 x^3-5x+4 ?
The tangent of y= 4/3 x^3-5x+4 is y=(4a_{0}^2-5)x-8/3 a_{0}^3+4