What is the tangent of f(x)=(4x+5)^4,\at x=-1 ?

The tangent of f(x)=(4x+5)^4,\at x=-1 is y=16x+17

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tangent of f(x)=(4x+5)^4,\at x=-1

tangent of

f (x) = (4 x + 5)^{4}, at x = - 1

y = 16 x + 17

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

f (x) = (4 x + 5)^{4} : \frac{df ( x )}{d x} = 16 (4 x + 5)^{3}

Compute the slope of

f (x) = (4 x + 5)^{4}

x = - 1 : m = 16

Find the line with slope m=

16

and passing through

(- 1, 1) : y = 16 x + 17

y = 16 x + 17

n = 1 \sum \infty \frac{n}{3 n + 1}

\frac{\partial}{\partial x} (\frac{1}{x} cos (\frac{y}{x}))

y^{^{'}} = (\frac{2 y + 3}{4 x + 5})^{2}

\int \frac{1}{( 25 x ^{2} + 4 ) ^{2}} d x

f (x) = (3 x^{3} + 9 x + 7) (x^{2} + 8)

What is the tangent of f(x)=(4x+5)^4,\at x=-1 ?
The tangent of f(x)=(4x+5)^4,\at x=-1 is y=16x+17