在同一坐标系中画出 and 的函数图像。清晰地标出这两个图像交点的坐标。 解不等式 。 因此求出 ,给出答案的精确形式。
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完整解法
1(a)
1. Sketch the graphs and find their points of intersection using a graphing calculator.
Using a graphing calculator, plot and .
- The graph of is an exponential curve with a horizontal asymptote at .
- The graph of is a straight line with an -intercept at and a -intercept at .
Equating both functions on the calculator yields the points of intersection:
- First intersection: exactly.
- Second intersection: to 3 significant figures.
2. Determine the solution to the inequality $3e^x > x + 3$ using the sketched graphs.
The inequality is satisfied in the regions where the graph of lies strictly above the line .
From the sketch, the exponential curve is above the straight line for -values less than the smaller root and greater than the larger root.
Thus, the solution is or .
Points of intersection: and
Solution to the inequality:
1(b)
1. Split the definite integral by analyzing the sign of the expression inside the modulus.
\text{From part (a), the graphs of } y = 3e^x \text{ and } y = x + 3 \text{ intersect at } x = 0.
\text{For } x \in [-2, 0], 3e^x \leq x + 3 \implies 3e^x - x - 3 \leq 0.
\text{For } x \in [0, 2], 3e^x \geq x + 3 \implies 3e^x - x - 3 \geq 0.
\int_{-2}^{2} |3e^x - x - 3| , \mathrm{d}x
= \int_{-2}^{0} -(3e^x - x - 3) , \mathrm{d}x + \int_{0}^{2} (3e^x - x - 3) , \mathrm{d}x
2. Integrate and evaluate the limits to find the exact value.
\begin{aligned}
&\int_{-2}^{0} -(3e^x - x - 3) , \mathrm{d}x + \int_{0}^{2} (3e^x - x - 3) , \mathrm{d}x \
&= -\left[ 3e^x - \frac{1}{2}x^2 - 3x \right]{-2}^{0} + \left[ 3e^x - \frac{1}{2}x^2 - 3x \right]{0}^{2} \
&= -\left( (3e^0 - 0 - 0) - \left(3e^{-2} - \frac{1}{2}(-2)^2 - 3(-2)\right) \right) + \left( \left(3e^2 - \frac{1}{2}(2)^2 - 3(2)\right) - (3e^0 - 0 - 0) \right) \
&= -\left( 3 - (3e^{-2} - 2 + 6) \right) + \left( (3e^2 - 2 - 6) - 3 \right) \
&= -(3 - 3e^{-2} - 4) + (3e^2 - 11) \
&= -(-1 - 3e^{-2}) + 3e^2 - 11 \
&= 1 + 3e^{-2} + 3e^2 - 11 \
&= 3e^2 + \frac{3}{e^2} - 10
\end{aligned}
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