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

The tangent of f(x)=x^4-17x^2+16 is y=(4a_{0}^3-34a_{0})x-3a_{0}^4+17a_{0}^2+16

tangent of f(x)=x^4-17x^2+16

tangent of

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

y = (4 a_{0}^{3} - 34 a_{0}) x - 3 a_{0}^{4} + 17 a_{0}^{2} + 16

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 17 a_{0}^{2} + 16)

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 = 4 a_{0}^{3} - 34 a_{0}

Find the line with slope m=

4 a_{0}^{3} - 34 a_{0}

and passing through

(a_{0}, a_{0}^{4} - 17 a_{0}^{2} + 16) : y = (4 a_{0}^{3} - 34 a_{0}) x - 3 a_{0}^{4} + 17 a_{0}^{2} + 16

y = (4 a_{0}^{3} - 34 a_{0}) x - 3 a_{0}^{4} + 17 a_{0}^{2} + 16

\int π (\frac{sin ( x )}{x})^{2} d x

\int_{- 1}^{1} \frac{5 x}{( x + 4 ) ^{2}} d x

y = ∣ x - 6 ∣, y = \frac{x}{2}

n = 1 \sum \infty \frac{2 n}{5 n - 3}

e^{s i n (2 x)}

What is the tangent of f(x)=x^4-17x^2+16 ?
The tangent of f(x)=x^4-17x^2+16 is y=(4a_{0}^3-34a_{0})x-3a_{0}^4+17a_{0}^2+16