Question 1171449: ``The value of function -10000e^(-0.03x)-120x+10000 at its critical points is
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! To find the value of the function at its critical points, we need to:
1. **Find the derivative of the function.**
2. **Set the derivative equal to zero and solve for x.** This will give us the critical points.
3. **Plug the critical points back into the original function to find the function's value.**
Let's do this step-by-step:
**1. Find the derivative:**
The given function is:
f(x) = -10000e^(-0.03x) - 120x + 10000
The derivative, f'(x), is:
f'(x) = -10000 * (-0.03) * e^(-0.03x) - 120
f'(x) = 300e^(-0.03x) - 120
**2. Set the derivative to zero and solve for x:**
300e^(-0.03x) - 120 = 0
300e^(-0.03x) = 120
e^(-0.03x) = 120 / 300
e^(-0.03x) = 0.4
Take the natural logarithm of both sides:
ln(e^(-0.03x)) = ln(0.4)
-0.03x = ln(0.4)
x = ln(0.4) / -0.03
x ≈ 30.543
**3. Plug the critical point back into the original function:**
f(x) = -10000e^(-0.03x) - 120x + 10000
f(30.543) = -10000e^(-0.03 * 30.543) - 120 * 30.543 + 10000
f(30.543) = -10000e^(-0.91629) - 3665.16 + 10000
f(30.543) = -10000 * 0.4 + 6334.84
f(30.543) = -4000 + 6334.84
f(30.543) = 2334.84
**Therefore, the value of the function at its critical point is approximately 2334.84.**
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