To determine [tex]\( F'(x) \)[/tex] for the function [tex]\( F(x) = \int_{x^2 - 1}^{0} f(t) \, dt \)[/tex], we will use the Fundamental Theorem of Calculus along with the Chain Rule. Let's break down the process step by step:
1. Express [tex]\(F(x)\)[/tex]:
[tex]\[
F(x) = \int_{x^2 - 1}^{0} f(t) \, dt
\][/tex]
2. Apply the Fundamental Theorem of Calculus:
The derivative of an integral with variable limits [tex]\(a(x)\)[/tex] to [tex]\(b(x)\)[/tex] is given by:
[tex]\[
\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t) \, dt \right) = f\big(b(x)\big) \cdot b'(x) - f\big(a(x)\big) \cdot a'(x)
\][/tex]
Here, [tex]\(a(x) = x^2 - 1\)[/tex] and [tex]\(b(x) = 0\)[/tex].
3. Differentiate the limits:
- The upper limit [tex]\(b(x)\)[/tex] is [tex]\(0\)[/tex], which is a constant. Therefore:
[tex]\[
b'(x) = \frac{d}{dx}(0) = 0
\][/tex]
- The lower limit [tex]\(a(x)\)[/tex] is [tex]\(x^2 - 1\)[/tex]. Therefore:
[tex]\[
a'(x) = \frac{d}{dx}(x^2 - 1) = 2x
\][/tex]
4. Substitute into the derivative formula:
Using the formula from the Fundamental Theorem of Calculus:
[tex]\[
F'(x) = f\big(0\big) \cdot 0 - f\big(x^2 - 1\big) \cdot 2x
\][/tex]
Since [tex]\(f(0) \cdot 0 = 0\)[/tex], the expression simplifies to:
[tex]\[
F'(x) = -f\big(x^2 - 1\big) \cdot 2x
\][/tex]
5. Evaluate [tex]\(F'(3)\)[/tex]:
- First, evaluate the argument of the function [tex]\(f\)[/tex]:
[tex]\[
x^2 - 1 = 3^2 - 1 = 9 - 1 = 8
\][/tex]
Thus:
[tex]\[
F'(3) = -f(8) \cdot 2 \cdot 3
\][/tex]
- We know from the problem statement that:
[tex]\[
f(8) = -4
\][/tex]
Substituting this value in:
[tex]\[
F'(3) = -(-4) \cdot 2 \cdot 3 = 4 \cdot 2 \cdot 3 = 24
\][/tex]
Therefore, [tex]\( F'(3) = 24 \)[/tex].
