Determine the equation of the line tangent to the graph of y=11e^(5x^2)-6x at x=3 in the form y= mx+b  

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The slope of the tangent to the graph of f(x) at x = a is equal to f'(a).

For the function `y = 11e^(5x^2)-6x, y' = 11*e^(5x^2)*10x - 6` .

At x = 3, `y' = 11*e^45*30 - 6 = 330*e^45 - 6 ~~ 1.15*10^22`

At x = 3, `y = 11*e^45 - 18 ~~ 3.84*10^20`

The equation of the tangent is `(y - 3.84*10^20)/(x - 3) = 1.15*10^22`

=> `y = 1.15*10^22*x - 3.45*10^22 + 3.84*10^20`

The required tangent is `y = 1.15*10^22*x - 3.41*10^22`

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