What is the tangent of f(x)=8x^2+5x,\at x=3 ?

The tangent of f(x)=8x^2+5x,\at x=3 is y=53x-72

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tangent of f(x)=8x^2+5x,\at x=3

tangent of

f (x) = 8 x^{2} + 5 x, at x = 3

y = 53 x - 72

Solution steps

Find the tangent point:

(3, 87)

Find the slope of

f (x) = 8 x^{2} + 5 x : \frac{df ( x )}{d x} = 16 x + 5

Compute the slope of

f (x) = 8 x^{2} + 5 x

x = 3 : m = 53

Find the line with slope m=

53

and passing through

(3, 87) : y = 53 x - 72

y = 53 x - 72

\frac{d}{d x} (cos^{3 (\frac{x ^{2}}{x - 1})} (x))

\frac{\partial}{\partial x} (\frac{y}{x ^{z}})

x \to 1.001 lim (x^{2} - 2 x - 1)

\int_{2}^{4} x^{3} d x

\int e^{x} 5 - e^{x} d x

What is the tangent of f(x)=8x^2+5x,\at x=3 ?
The tangent of f(x)=8x^2+5x,\at x=3 is y=53x-72