What is the tangent of f(x)= 1/3 x^3-5x^2+26x+23 ?

The tangent of f(x)= 1/3 x^3-5x^2+26x+23 is y=(a_{0}^2-10a_{0}+26)x-2/3 a_{0}^3+5a_{0}^2+23

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tangent of f(x)= 1/3 x^3-5x^2+26x+23

tangent of

f (x) = \frac{1}{3} x^{3} - 5 x^{2} + 26 x + 23

y = (a_{0}^{2} - 10 a_{0} + 26) x - \frac{2}{3} a_{0}^{3} + 5 a_{0}^{2} + 23

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{3} a_{0}^{3} - 5 a_{0}^{2} + 26 a_{0} + 23)

Find the slope of

f (x) = \frac{1}{3} x^{3} - 5 x^{2} + 26 x + 23 : \frac{df ( x )}{d x} = x^{2} - 10 x + 26

Compute the slope of

f (x) = \frac{1}{3} x^{3} - 5 x^{2} + 26 x + 23

x = a_{0} : m = a_{0}^{2} - 10 a_{0} + 26

Find the line with slope m=

a_{0}^{2} - 10 a_{0} + 26

and passing through

(a_{0}, \frac{1}{3} a_{0}^{3} - 5 a_{0}^{2} + 26 a_{0} + 23) : y = (a_{0}^{2} - 10 a_{0} + 26) x - \frac{2}{3} a_{0}^{3} + 5 a_{0}^{2} + 23

y = (a_{0}^{2} - 10 a_{0} + 26) x - \frac{2}{3} a_{0}^{3} + 5 a_{0}^{2} + 23

\int y^{2} - x^{2} d x

f (x) = \frac{1}{x ^{2}} + 5 \frac{1}{x ^{3}}

\frac{\partial}{\partial t} (t e^{t} + 2)

f (x) = 3 x^{2} - 5 x + 6

\frac{d}{d x} ((e^{x})^{c o s (x)})

What is the tangent of f(x)= 1/3 x^3-5x^2+26x+23 ?
The tangent of f(x)= 1/3 x^3-5x^2+26x+23 is y=(a_{0}^2-10a_{0}+26)x-2/3 a_{0}^3+5a_{0}^2+23