What is the tangent of f(x)=(2x+3)^5,\at x=-1 ?

The tangent of f(x)=(2x+3)^5,\at x=-1 is y=10x+11

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tangent of f(x)=(2x+3)^5,\at x=-1

tangent of

f (x) = (2 x + 3)^{5}, at x = - 1

y = 10 x + 11

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

f (x) = (2 x + 3)^{5} : \frac{df ( x )}{d x} = 10 (2 x + 3)^{4}

Compute the slope of

f (x) = (2 x + 3)^{5}

x = - 1 : m = 10

Find the line with slope m=

10

and passing through

(- 1, 1) : y = 10 x + 11

y = 10 x + 11

\frac{d y}{d x} - 2 y = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 x + 7

\int sin^{3} (x co) s^{7} x d x

(cos (6 x)) y^{^{'}} + 6 (tan (6 x)) y = - sec (6 x)

What is the tangent of f(x)=(2x+3)^5,\at x=-1 ?
The tangent of f(x)=(2x+3)^5,\at x=-1 is y=10x+11