The function f whose graph passes through (4, - 20) and whose derivative is `cos (sqrt(4-x))` is given by

Options

A. `sqrt(4-x)sinsqrt(4-x)+(4-6x)cossqrt(4-x)`
B. `sqrt(4-x)sinsqrt(4-x)+sqrt(4-x)cossqrt(4-x)-20`
C. `-2(sqrt(4-x)sinsqrt(4-x)+cossqrt(4-x))-18`
D. `-(2x+12)sqrt(4-x)sinsqrt(4-x)+(4-6x)cossqrt(4-x)`

Answer

To find the function \( f \) whose graph passes through the point \( (4, -20) \) and whose derivative is \( f'(x) = \cos(\sqrt{4 - x}) \), we will follow these steps:

### Step 1: Set up the integral
We know that \( f'(x) = \cos(\sqrt{4 - x}) \). To find \( f(x) \), we need to integrate \( f'(x) \):
\[ f(x) = \int \cos(\sqrt{4 - x}) \, dx \]

### Step 2: Use substitution
Let \( t = \sqrt{4 - x} \). Then, we can express \( x \) in terms of \( t \):
\[ x = 4 - t^2 \]
Now, differentiate both sides with respect to \( x \):
\[ dx = -2t \, dt \]

### Step 3: Substitute in the integral
Substituting \( t \) and \( dx \) into the integral, we have:
\[ f(x) = \int \cos(t) (-2t) \, dt = -2 \int t \cos(t) \, dt \]

### Step 4: Apply integration by parts
Using integration by parts, where we let:
- \( u = t \) (thus \( du = dt \))
- \( dv = \cos(t) \, dt \) (thus \( v = \sin(t) \))

Now, apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
This gives us:
\[ -2 \left( t \sin(t) - \int \sin(t) \, dt \right) \]
Calculating the integral of \( \sin(t) \):
\[ \int \sin(t) \, dt = -\cos(t) \]
Thus,
\[ f(x) = -2 \left( t \sin(t) + \cos(t) \right) + C \]

### Step 5: Substitute back for \( t \)
Now, replace \( t \) back with \( \sqrt{4 - x} \):
\[ f(x) = -2 \left( \sqrt{4 - x} \sin(\sqrt{4 - x}) + \cos(\sqrt{4 - x}) \right) + C \]

### Step 6: Use the point to find \( C \)
We know that \( f(4) = -20 \):
\[ f(4) = -2 \left( \sqrt{4 - 4} \sin(\sqrt{4 - 4}) + \cos(\sqrt{4 - 4}) \right) + C = -20 \]
This simplifies to:
\[ -2(0 + 1) + C = -20 \]
\[ -2 + C = -20 \]
\[ C = -20 + 2 = -18 \]

### Step 7: Write the final function
Substituting \( C \) back into the equation, we get:
\[ f(x) = -2 \left( \sqrt{4 - x} \sin(\sqrt{4 - x}) + \cos(\sqrt{4 - x}) \right) - 18 \]

Thus, the function \( f \) is:
\[ f(x) = -2 \sqrt{4 - x} \sin(\sqrt{4 - x}) + \cos(\sqrt{4 - x}) - 18 \]