What is the tangent of f(x)=36x-16x^2,\at x=1 ?

The tangent of f(x)=36x-16x^2,\at x=1 is y=4x+16

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tangent of f(x)=36x-16x^2,\at x=1

tangent of

f (x) = 36 x - 16 x^{2}, at x = 1

y = 4 x + 16

Solution steps

Find the tangent point:

(1, 20)

Find the slope of

f (x) = 36 x - 16 x^{2} : \frac{df ( x )}{d x} = 36 - 32 x

Compute the slope of

f (x) = 36 x - 16 x^{2}

x = 1 : m = 4

Find the line with slope m=

4

and passing through

(1, 20) : y = 4 x + 16

y = 4 x + 16

\frac{d y}{d x} = \frac{3 x ^{2}}{x ^{3} + y + 1}

\int_{0}^{1} (3 x^{2}) d x

f (x) = e^{x^{2}}

n \to \infty lim ((- 1)^{n})

\frac{d}{d x} (\frac{x ^{3}}{3 - 6 x ^{2} + 8 x - 7})

What is the tangent of f(x)=36x-16x^2,\at x=1 ?
The tangent of f(x)=36x-16x^2,\at x=1 is y=4x+16