Find 
g(x) if g'(x) = (60x2 − 80x) · 74x3 − 8x2 and g(0) = 1.
g(x) =

To find \( g(x) \), given \( g'(x) = (60x^2 - 80x) \cdot (74x^3 - 8x^2) \) and \( g(0) = 1 \), we need to integrate \( g'(x) \) with respect to \( x \) and then apply the initial condition \( g(0) = 1 \).

### Step 1: Simplify \( g'(x) \)
Rewrite \( g'(x) \):
\[
g'(x) = (60x^2 - 80x)(74x^3 - 8x^2)
\]

Expand \( g'(x) \) by multiplying the terms:
\[
g'(x) = (60x^2 \cdot 74x^3) - (60x^2 \cdot 8x^2) - (80x \cdot 74x^3) + (80x \cdot 8x^2)
\]

This gives:
\[
g'(x) = 4440x^5 - 480x^4 - 5920x^4 + 640x^3
\]
Combine like terms:
\[
g'(x) = 4440x^5 - 6400x^4 + 640x^3
\]

### Step 2: Integrate \( g'(x) \) to find \( g(x) \)
Now, we integrate \( g'(x) \) with respect to \( x \):
\[
g(x) = \int (4440x^5 - 6400x^4 + 640x^3) \, dx
\]

Integrating each term separately:
1. \(\int 4440x^5 \, dx = \frac{4440}{6}x^6 = 740x^6\)
2. \(\int -6400x^4 \, dx = -\frac{6400}{5}x^5 = -1280x^5\)
3. \(\int 640x^3 \, dx = \frac{640}{4}x^4 = 160x^4\)

Thus,
\[
g(x) = 740x^6 - 1280x^5 + 160x^4 + C
\]

### Step 3: Apply the initial condition \( g(0) = 1 \)
Substitute \( x = 0 \) and \( g(0) = 1 \):
\[
g(0) = 740 \cdot 0^6 - 1280 \cdot 0^5 + 160 \cdot 0^4 + C = 1
\]
This simplifies to \( C = 1 \).

### Final Answer
Therefore,
\[
g(x) = 740x^6 - 1280x^5 + 160x^4 + 1
\]

Would you like more details on any part of the solution? 

Here are five related questions to consider:
1. How would the solution change if \( g(0) \) had a different value?
2. How can you check the correctness of \( g(x) \) by differentiating it?
3. What techniques can be used to simplify products of polynomials before integrating?
4. How would the solution differ if there were additional terms in \( g'(x) \)?
5. How does the degree of \( g(x) \) relate to the degree of \( g'(x) \)?

**Tip:** Always remember to apply initial conditions after integrating to determine the constant of integration accurately.

Find g(x) if g'(x) = (60x^2 − 80x) · 74x^3 − 8x^2 and g(0) = 1.

Learn how to find the function g(x) from its derivative g'(x) = (60x^2 − 80x)(74x^3 − 8x^2) with initial condition g(0) = 1. This step-by-step solution covers the process of simplifying, integrating, and applying initial conditions to obtain the final function. Suitable for students in Grades 11-12.

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