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Compute f'(a). f(x) = x^2 + 11x, when a = 6
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f ' (a)
find f'(x) = 2x + 11 then substitute x = 6 into the derivative to find f'(a)
12+11 = 23
did you all use the power rule? and what about this... Find f(3) and f'(3), assuming that the tangent line to y = f(x) at a = 3 has equation y = 6x + 8
how would you find the tangent line equation if you knew f '(X) or just f(x)
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f'(x) at a point x = 6 will give the gradient of that tangent at x = 6..... sub x = 6 into the original equation to find the y value... then you have the gradient and point... which can be used the the point slope formula y - f(a) = f'(a)(x - a) to get the equation of the tangent
so what is a?
in your example where x = 3 then y - f(3) = f'(3)(x-3)
so 26 - 26 = 6 - 3?
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