What is the tangent of f(x)=8x^3-24x^2-193x+48 ?

The tangent of f(x)=8x^3-24x^2-193x+48 is y=(24a_{0}^2-48a_{0}-193)x-16a_{0}^3+24a_{0}^2+48

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tangent of f(x)=8x^3-24x^2-193x+48

tangent of

f (x) = 8 x^{3} - 24 x^{2} - 193 x + 48

y = (24 a_{0}^{2} - 48 a_{0} - 193) x - 16 a_{0}^{3} + 24 a_{0}^{2} + 48

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{3} - 24 a_{0}^{2} - 193 a_{0} + 48)

Find the slope of

f (x) = 8 x^{3} - 24 x^{2} - 193 x + 48 : \frac{df ( x )}{d x} = 24 x^{2} - 48 x - 193

Compute the slope of

f (x) = 8 x^{3} - 24 x^{2} - 193 x + 48

x = a_{0} : m = 24 a_{0}^{2} - 48 a_{0} - 193

Find the line with slope m=

24 a_{0}^{2} - 48 a_{0} - 193

and passing through

(a_{0}, 8 a_{0}^{3} - 24 a_{0}^{2} - 193 a_{0} + 48) : y = (24 a_{0}^{2} - 48 a_{0} - 193) x - 16 a_{0}^{3} + 24 a_{0}^{2} + 48

y = (24 a_{0}^{2} - 48 a_{0} - 193) x - 16 a_{0}^{3} + 24 a_{0}^{2} + 48

\frac{d}{d x} (sin (x) (x^{2} - 27 x + 6))

\frac{d}{d x} (ln (1 + x))

n = 1 \sum \infty (sin (9))^{n}

49 y^{^{''}} + 14 y^{^{'}} + y = 0

\frac{d}{d x} ((2 x^{2} + 1)^{\frac{3}{2}})

What is the tangent of f(x)=8x^3-24x^2-193x+48 ?
The tangent of f(x)=8x^3-24x^2-193x+48 is y=(24a_{0}^2-48a_{0}-193)x-16a_{0}^3+24a_{0}^2+48