Keeping in mind that for a cost C(x) and profit P(x) and revenue R(x), the marginal cost, marginal profit and marginal revenue are respectively dC/dx, dP/dx and dR/dx, then
![\bf P(x)=0.03x^2-3x+3x^{0.8}-4400 \\\\\\ \stackrel{marginal~profit}{\cfrac{dP}{dx}}=0.06x-3+2.4x^{-0.2} \\\\\\ \cfrac{dP}{dx}=0.06x-3+2.4\cdot \cfrac{1}{x^{0.2}}\implies \cfrac{dP}{dx}=0.06x-3+2.4\cdot \cfrac{1}{x^{\frac{1}{5}}} \\\\\\ \cfrac{dP}{dx}=0.06x-3+\cfrac{2.4}{\sqrt[5]{x}}](https://tex.z-dn.net/?f=%5Cbf%20P%28x%29%3D0.03x%5E2-3x%2B3x%5E%7B0.8%7D-4400%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bmarginal~profit%7D%7B%5Ccfrac%7BdP%7D%7Bdx%7D%7D%3D0.06x-3%2B2.4x%5E%7B-0.2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B2.4%5Ccdot%20%5Ccfrac%7B1%7D%7Bx%5E%7B0.2%7D%7D%5Cimplies%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B2.4%5Ccdot%20%5Ccfrac%7B1%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7Bx%7D%7D)
Start by making the denominators equal.
1/3 becomes 4/12
1/6 becomes 2/12
and you have 5/12.
Add together to get 10/12.
Once simplified, the answer should be 5/6.
Answer:
The Exponential function is

And the polynomial function is

And, we have to find the value of x for which, exponential function exceeds the polynomial function which can be written as

1. When , x= -1
LHS

RHS

2. When , x=0
L HS

RHS

3. When ,x= 0.5
L HS

RHS

4. When , x=2
LHS

RHS

The Minimum value for which exponential function exceeds the polynomial function is , x= 0.5
But,there is other value for which exponential function exceeds the polynomial function is , x=2.
Oh, you're from connexus right? Well the answer is ounces