B. Find the equation of a tangent line to the curve y = e(x2) that also passes through the point (1,0). You can approximate your final answer with your calculator after you've shown your work.
My work:
set the slope of the tangent line delta(y)/delta(x) equal to the derivative of ex^2 at x = a
(e^(x^2))’ = 2xe^(x^2)
The slope of the tangent line is delta(y)/delta(x) = ((e^(a^2)) – 0)/(a – 1)
2ae^(a^2) = (e(a^2))/(a – 1)
2a = 1/(a – 1)
I can't figure out how to get a all by itself. I can use a calculator to solve for it, but I don't want a simplified answer.
My work:
set the slope of the tangent line delta(y)/delta(x) equal to the derivative of ex^2 at x = a
(e^(x^2))’ = 2xe^(x^2)
The slope of the tangent line is delta(y)/delta(x) = ((e^(a^2)) – 0)/(a – 1)
2ae^(a^2) = (e(a^2))/(a – 1)
2a = 1/(a – 1)
I can't figure out how to get a all by itself. I can use a calculator to solve for it, but I don't want a simplified answer.
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You're doing fine so far. Now you have to solve 2a(a-1) = 1 ---> 2*a^2 - 2a - 1 = 0 using the quadratic equation, which gives
a = (2 +/- sqrt(4 + 8)) / 4 = (2 +/- 2sqrt(3)) / 4 = (1 +/- sqrt(3)) / 2, so
a = 1.366, y(a) = 6.462 or a = -0.366, y(a) = 1.143 so you have two possible tangent lines fitting the requirements.
In the first case you have slope (6.462 - 0)/(1.366 - 1) = 17.66 and y-int. b = - 17.66,
and in the second case you have slope (1.143 - 0)/(-0.366 - 1) = -0.837 and b = 0.837.
a = (2 +/- sqrt(4 + 8)) / 4 = (2 +/- 2sqrt(3)) / 4 = (1 +/- sqrt(3)) / 2, so
a = 1.366, y(a) = 6.462 or a = -0.366, y(a) = 1.143 so you have two possible tangent lines fitting the requirements.
In the first case you have slope (6.462 - 0)/(1.366 - 1) = 17.66 and y-int. b = - 17.66,
and in the second case you have slope (1.143 - 0)/(-0.366 - 1) = -0.837 and b = 0.837.