What is the tangent of f(x)=37x-16x^2,\at x=1 ?

The tangent of f(x)=37x-16x^2,\at x=1 is y=5x+16

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tangent of f(x)=37x-16x^2,\at x=1

tangent of

f (x) = 37 x - 16 x^{2}, at x = 1

y = 5 x + 16

Solution steps

Find the tangent point:

(1, 21)

Find the slope of

f (x) = 37 x - 16 x^{2} : \frac{df ( x )}{d x} = 37 - 32 x

Compute the slope of

f (x) = 37 x - 16 x^{2}

x = 1 : m = 5

Find the line with slope m=

5

and passing through

(1, 21) : y = 5 x + 16

y = 5 x + 16

\int (2 sec^{2} (x)) d x

\int \frac{6}{x ( x ^{2} + 4 ) ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (e^{x} cos (x))

x \frac{d y}{d x} + (1 + x) y = \frac{sin ( 2 x )}{e ^{x}}

\frac{x d y}{d x} + 4 y = x^{3} - x

What is the tangent of f(x)=37x-16x^2,\at x=1 ?
The tangent of f(x)=37x-16x^2,\at x=1 is y=5x+16