What is the tangent of f(x)=(4x+3)(-x+1),\at x=4 ?

The tangent of f(x)=(4x+3)(-x+1),\at x=4 is y=-31x+67

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tangent of f(x)=(4x+3)(-x+1),\at x=4

tangent of

f (x) = (4 x + 3) (- x + 1), at x = 4

y = - 31 x + 67

Solution steps

Find the tangent point:

(4, - 57)

Find the slope of

f (x) = (4 x + 3) (- x + 1) : \frac{df ( x )}{d x} = - 8 x + 1

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 = - 31

Find the line with slope m=

- 31

and passing through

(4, - 57) : y = - 31 x + 67

y = - 31 x + 67

\frac{d}{d x} (\frac{6}{4 x})

f (x) = 7 x - x^{2}, (1, 6)

\int_{- 1}^{1} \frac{1}{x} d x

y = \frac{x ^{2} + 5}{6 x - x ^{2}}, (1, 5)

\int - 7 lo g_{2} (x) d x

What is the tangent of f(x)=(4x+3)(-x+1),\at x=4 ?
The tangent of f(x)=(4x+3)(-x+1),\at x=4 is y=-31x+67