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

The tangent of f(x)=x^4-5x^3+2,\at x=2 is y=-28x+34

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tangent of f(x)=x^4-5x^3+2,\at x=2

tangent of

f (x) = x^{4} - 5 x^{3} + 2, at x = 2

y = - 28 x + 34

Solution steps

Find the tangent point:

(2, - 22)

Find the slope of

f (x) = x^{4} - 5 x^{3} + 2 : \frac{df ( x )}{d x} = 4 x^{3} - 15 x^{2}

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 = - 28

Find the line with slope m=

- 28

and passing through

(2, - 22) : y = - 28 x + 34

y = - 28 x + 34

\frac{d}{d x} (5 ln (x^{3}))

n = 0 \sum \infty \frac{1}{( 2 n ! )}

\int x cos (mπ x) d x

\frac{\partial}{\partial x} (9 x e^{6 x y})

e^{5 (t)}

What is the tangent of f(x)=x^4-5x^3+2,\at x=2 ?
The tangent of f(x)=x^4-5x^3+2,\at x=2 is y=-28x+34