What is the tangent of f(x)=(x-4)(x^2+8),\at x=1 ?

The tangent of f(x)=(x-4)(x^2+8),\at x=1 is y=3x-30

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tangent of f(x)=(x-4)(x^2+8),\at x=1

tangent of

f (x) = (x - 4) (x^{2} + 8), at x = 1

y = 3 x - 30

Solution steps

Find the tangent point:

(1, - 27)

Find the slope of

f (x) = (x - 4) (x^{2} + 8) : \frac{df ( x )}{d x} = 3 x^{2} - 8 x + 8

Compute the slope of

f (x) = (x - 4) (x^{2} + 8)

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, - 27) : y = 3 x - 30

y = 3 x - 30

(8 x + 9) (9 x - 10)

\frac{d}{d x} (e^{- c o s (x)})

\frac{e ^{x}}{5 x + 3}

\int \frac{1}{( 4 x ^{2} - 9 ) ^{\frac{3}{2}}} d x

\int x^{2} sin (x^{3}) d x

What is the tangent of f(x)=(x-4)(x^2+8),\at x=1 ?
The tangent of f(x)=(x-4)(x^2+8),\at x=1 is y=3x-30