JustAnswer
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|(a) y = (x^2)(e^x) |
|Differentiating with respect to x using product rule, we have |
|dy/dx = (x^2)(e^x) + (e^x)(2x) |
|= (e^x)(x^2 + 2x) |
|= x(x + 2)(e^x) |
|(b) y = (e^x + 2)^(3/2) |
|Differentiating with respect to x using chain rule, we have |
|dy/dx = (3/2)(e^x + 2)^(3/2 - 1) * (d/dx)(e^x + 2) |
|= (3/2)(e^x + 2)^(1/2) * (e^x) |
|= (3/2)(e^x) √(e^x + 2) |
|(c) y = e^(-3x) |
|Differentiating with respect to x using chain rule, we have |
|dy/dx = [e^(-3x)] * (d/dx)(-3x) |
|= -3 e^(-3x) |
|(d) y = (1/2)(e^x - e^-x) |
|Differentiating with respect to x using chain rule, we have |
|dy/dx = (1/2)[e^x - (e^-x) * (d/dx)(-x)] |
|= (1/2)(e^x + e^-x) |
[pic]
|Graph of P(t) = 300000 e^(-0.09t + (√t /2)) for 0 ≤ t ≤ 10 |
|t |
|P(t) = 300000 e^(-0.09t + (√t /2)) |
| |
|0 |
|300000 |
| |
|1 |
|452045.34 |
| |
|2 |
|508207.21 |
| |
|3 |
|544467.30 |
| |
|4 |
|568944.26 |
| |
|5 |
|585119.71 |
| |
|6 |
|594979.73 |
| |
|7 |
|599837.11 |
| |
|8 |
|600640.17 |
| |
|9 |
|598114.66 |
| |
|10 |
|592838.08 |
| |
| |
|[pic] |
|The tangent line to the graph is horizontal (slope = 0) at x = 7.7 approximately. |
|Now when t = 7.7, P(t) = 300000 e^(-0.09*7.7 + (√7.7 /2)) = 600779 |
|( The optimum value of the building is $600779 and happens 7.7 years later. |
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|f(x) = x e^(-x) |
|Differentiating with respect to x using product rule and chain rule, we have |
|f’(x) = x * e^(-x) * (d/dx)(-x) + e^(-x) = [e^(-x)](1 - x) |
|f’(0) = [e^(-0)](1 - 0) = 1 |
|f(0) = 0 |
|The equation of the line tangent at x = 0 is y - 0 = 1(x - 0), that is y = x |
|When x = 0, f(x) = 0, and the graph passes through the origin. Also, it is tangential to the line y = x at the origin. |
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|(a) R(x) = x * p(x) = x * 100e^(-0.0001x) |
|R(x) = 100x e^(-0.0001x) |
|(b) R’(x) = 100 [x * (-0.0001)e^(-0.0001x) + e^(-0.0001x)] |
|R’(x) = 100 e^(-0.0001x) * [1 - 0.0001x] |
|(c) When x = 10, R’(10) = 100 e^(-0.0001*10) * [1 - 0.0001*10] |
|R’(10) = 99.80 |
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