What is the tangent of f(x)=x^4-13^2+36,\at x=3 ?

The tangent of f(x)=x^4-13^2+36,\at x=3 is y=108x-376

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tangent of f(x)=x^4-13^2+36,\at x=3

tangent of

f (x) = x^{4} - 1 3^{2} + 36, at x = 3

y = 108 x - 376

Solution steps

Find the tangent point:

(3, - 52)

Find the slope of

f (x) = x^{4} - 1 3^{2} + 36 : \frac{df ( x )}{d x} = 4 x^{3}

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 = 108

Find the line with slope m=

108

and passing through

(3, - 52) : y = 108 x - 376

y = 108 x - 376

x \to 1 lim (\frac{3 x - 2}{x + 2})

x \to 4 - lim (x^{2} + 1)

y^{^{'}} = \frac{x y ^{3}}{1 + x ^{2}}

\int \frac{e ^{4 x}}{e ^{4 x} + 6} d x

n = 1 \sum \infty \frac{5}{2 ^{n}}

What is the tangent of f(x)=x^4-13^2+36,\at x=3 ?
The tangent of f(x)=x^4-13^2+36,\at x=3 is y=108x-376