What is the tangent of f(x)=x^4-17x^2+16,\at x=1 ?

The tangent of f(x)=x^4-17x^2+16,\at x=1 is y=-30x+30

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tangent of f(x)=x^4-17x^2+16,\at x=1

tangent of

f (x) = x^{4} - 17 x^{2} + 16, at x = 1

y = - 30 x + 30

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x^{4} - 17 x^{2} + 16 : \frac{df ( x )}{d x} = 4 x^{3} - 34 x

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 = - 30

Find the line with slope m=

- 30

and passing through

(1, 0) : y = - 30 x + 30

y = - 30 x + 30

sin^{2} (x), sin^{3} (x), [0, π]

y = 2 x^{3}, x = 0, x = 2

y = x^{3} - 13 x^{2} + 36 x, y = - x^{3} + 13 x^{2} - 36 x

\frac{d}{d x} (ln ∣ ln (3 x) ∣)

\frac{d}{d x} (e^{2 x - 4})

What is the tangent of f(x)=x^4-17x^2+16,\at x=1 ?
The tangent of f(x)=x^4-17x^2+16,\at x=1 is y=-30x+30