What is the tangent of f(x)=4x^3-7x^2-15x+10 ?

The tangent of f(x)=4x^3-7x^2-15x+10 is y=(12a_{0}^2-14a_{0}-15)x-8a_{0}^3+7a_{0}^2+10

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tangent of f(x)=4x^3-7x^2-15x+10

tangent of

f (x) = 4 x^{3} - 7 x^{2} - 15 x + 10

y = (12 a_{0}^{2} - 14 a_{0} - 15) x - 8 a_{0}^{3} + 7 a_{0}^{2} + 10

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{3} - 7 a_{0}^{2} - 15 a_{0} + 10)

Find the slope of

f (x) = 4 x^{3} - 7 x^{2} - 15 x + 10 : \frac{df ( x )}{d x} = 12 x^{2} - 14 x - 15

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 = 12 a_{0}^{2} - 14 a_{0} - 15

Find the line with slope m=

12 a_{0}^{2} - 14 a_{0} - 15

and passing through

(a_{0}, 4 a_{0}^{3} - 7 a_{0}^{2} - 15 a_{0} + 10) : y = (12 a_{0}^{2} - 14 a_{0} - 15) x - 8 a_{0}^{3} + 7 a_{0}^{2} + 10

y = (12 a_{0}^{2} - 14 a_{0} - 15) x - 8 a_{0}^{3} + 7 a_{0}^{2} + 10

y = 7 x - 4 x, (1, 3)

(\frac{y}{x} + 5 x) d x + (ln (x) - 2) d y = 0

0.08 e^{x} x^{2}

\frac{d}{d x} (20 x)

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

What is the tangent of f(x)=4x^3-7x^2-15x+10 ?
The tangent of f(x)=4x^3-7x^2-15x+10 is y=(12a_{0}^2-14a_{0}-15)x-8a_{0}^3+7a_{0}^2+10