What is the tangent of f(x)= 3/4 x^4-4/3 x^3+5/2 ?

The tangent of f(x)= 3/4 x^4-4/3 x^3+5/2 is y=(3a_{0}^3-4a_{0}^2)x-9/4 a_{0}^4+8/3 a_{0}^3+5/2

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tangent of f(x)= 3/4 x^4-4/3 x^3+5/2

tangent of

f (x) = \frac{3}{4} x^{4} - \frac{4}{3} x^{3} + \frac{5}{2}

y = (3 a_{0}^{3} - 4 a_{0}^{2}) x - \frac{9}{4} a_{0}^{4} + \frac{8}{3} a_{0}^{3} + \frac{5}{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{3}{4} a_{0}^{4} - \frac{4}{3} a_{0}^{3} + \frac{5}{2})

Find the slope of

f (x) = \frac{3}{4} x^{4} - \frac{4}{3} x^{3} + \frac{5}{2} : \frac{df ( x )}{d x} = 3 x^{3} - 4 x^{2}

Compute the slope of

f (x) = \frac{3}{4} x^{4} - \frac{4}{3} x^{3} + \frac{5}{2}

x = a_{0} : m = 3 a_{0}^{3} - 4 a_{0}^{2}

Find the line with slope m=

3 a_{0}^{3} - 4 a_{0}^{2}

and passing through

(a_{0}, \frac{3}{4} a_{0}^{4} - \frac{4}{3} a_{0}^{3} + \frac{5}{2}) : y = (3 a_{0}^{3} - 4 a_{0}^{2}) x - \frac{9}{4} a_{0}^{4} + \frac{8}{3} a_{0}^{3} + \frac{5}{2}

y = (3 a_{0}^{3} - 4 a_{0}^{2}) x - \frac{9}{4} a_{0}^{4} + \frac{8}{3} a_{0}^{3} + \frac{5}{2}

- 2 x + 5

(- 43, 4)

y = 3

y = sin (x) cos (x)

(2, 270)

What is the tangent of f(x)= 3/4 x^4-4/3 x^3+5/2 ?
The tangent of f(x)= 3/4 x^4-4/3 x^3+5/2 is y=(3a_{0}^3-4a_{0}^2)x-9/4 a_{0}^4+8/3 a_{0}^3+5/2