What is the tangent of f(x)=(x^2-48)^4,\at x=7 ?

The tangent of f(x)=(x^2-48)^4,\at x=7 is y=56x-391

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tangent of f(x)=(x^2-48)^4,\at x=7

tangent of

f (x) = (x^{2} - 48)^{4}, at x = 7

y = 56 x - 391

Solution steps

Find the tangent point:

(7, 1)

Find the slope of

f (x) = (x^{2} - 48)^{4} : \frac{df ( x )}{d x} = 8 x (x^{2} - 48)^{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 = 56

Find the line with slope m=

56

and passing through

(7, 1) : y = 56 x - 391

y = 56 x - 391

What is the tangent of f(x)=(x^2-48)^4,\at x=7 ?
The tangent of f(x)=(x^2-48)^4,\at x=7 is y=56x-391