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

The tangent of f(x)=5x^2-2x,\at x=3 is y=28x-45

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

tangent of

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

y = 28 x - 45

Solution steps

Find the tangent point:

(3, 39)

Find the slope of

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

Compute the slope of

f (x) = 5 x^{2} - 2 x

x = 3 : m = 28

Find the line with slope m=

28

and passing through

(3, 39) : y = 28 x - 45

y = 28 x - 45

f (x) = \frac{1}{2} x^{3} - 2 x + 3

\int \frac{12 x ^{3} - 24 x ^{2} + 7}{x ^{2} - 2 x} d x

\frac{s - 3}{s + 7}

x \to - 2 lim (\frac{- 1}{- x ^{2} + 4})

\frac{d y}{d x} = 4 x + \frac{9 x ^{2}}{( 3 x ^{3} + 1 ) ^{\frac{3}{2}}}

What is the tangent of f(x)=5x^2-2x,\at x=3 ?
The tangent of f(x)=5x^2-2x,\at x=3 is y=28x-45