Find 

f 

such that 

f prime left parenthesis x right parenthesisf′(x)equals=6 x minus 56x−5​, 

f left parenthesis 4 right parenthesisf(4)equals=0.

To find the function \( f(x) \), given that \( f'(x) = 6x - 5 \) and \( f(4) = 0 \), we can follow these steps:

### Step 1: Integrate \( f'(x) = 6x - 5 \)
To find \( f(x) \), we integrate \( f'(x) \):

\[
f(x) = \int (6x - 5) \, dx
\]

The integral of \( 6x \) is \( 3x^2 \), and the integral of \( -5 \) is \( -5x \). Therefore, we get:

\[
f(x) = 3x^2 - 5x + C
\]

where \( C \) is the constant of integration.

### Step 2: Use the initial condition \( f(4) = 0 \)
We are given that \( f(4) = 0 \), so we substitute \( x = 4 \) into the equation \( f(x) = 3x^2 - 5x + C \):

\[
0 = 3(4)^2 - 5(4) + C
\]

Simplifying:

\[
0 = 3(16) - 5(4) + C
\]
\[
0 = 48 - 20 + C
\]
\[
0 = 28 + C
\]
\[
C = -28
\]

### Step 3: Write the final function
Now that we have the value of \( C \), the function \( f(x) \) is:

\[
f(x) = 3x^2 - 5x - 28
\]

### Final answer:
\[
f(x) = 3x^2 - 5x - 28
\]

Would you like a more detailed explanation or any clarification?

Here are 5 related questions to expand your understanding:
1. How do you integrate a polynomial function like \( 6x - 5 \)?
2. What is the significance of the constant of integration in indefinite integrals?
3. How do you verify that \( f(x) = 3x^2 - 5x - 28 \) satisfies the given conditions?
4. What is the geometric interpretation of the derivative \( f'(x) = 6x - 5 \)?
5. How would you solve a similar problem if \( f'(x) \) involved a more complex function?

**Tip**: Always check the initial condition after solving for the constant to ensure your solution is correct.

Find f such that f′(x) = 6x − 5, f(4) = 0.

Learn how to solve the calculus problem of finding the function f(x) when given its derivative f′(x) = 6x − 5 and the initial condition f(4) = 0. This detailed solution covers integration and applying the initial value to find the constant of integration. Suitable for students in Grades 11-12.

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