hi everyone find equation of tangent line to curve y = 3x^2 + e^(1-x) when x =1 please show working so i can follow along thankyou!
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Originally Posted by tankertert hi everyone find equation of tangent line to curve y = 3x^2 + e^(1-x) when x =1 please show working so i can follow along thankyou! $\displaystyle f'(x)=6x-e^{1-x}$ , see WA Equation of tangent to curve at point $\displaystyle a$ is given by : $\displaystyle y=f(a)+f'(a)(x-a)$
so i should be subbing it in to get y = (3 x 1^2 + e^{1-1}) + (6 x 1 - e^{1-1})(x-1) and then solving for x?
simplify expression...
sorry thats i meant so answer is y = 3x +(x-1)(6x-1) +1 ?
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