Let's solve the problem step by step.
Step 1: Integrate [tex]\( f'(x) = a x^2 + b \)[/tex] to find [tex]\( f(x) \)[/tex].
Given [tex]\( f'(x) = a x^2 + b \)[/tex], we integrate with respect to [tex]\( x \)[/tex]:
[tex]\[
f(x) = \int (a x^2 + b) \, dx = \frac{a}{3} x^3 + b x + C
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
Step 2: Use the known gradient at point (3,5).
We know the gradient of [tex]\( f(x) \)[/tex] at [tex]\( x = 3 \)[/tex] is 4:
[tex]\[
f'(3) = 4
\][/tex]
Plug [tex]\( x = 3 \)[/tex] into the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[
a(3^2) + b = 4 \implies 9a + b = 4
\][/tex]
From this equation, we can solve for [tex]\( a \)[/tex] in terms of [tex]\( b \)[/tex]:
[tex]\[
9a + b = 4 \implies a = \frac{4 - b}{9}
\][/tex]
Step 3: Substitute [tex]\( a \)[/tex] back into [tex]\( f(x) \)[/tex].
Replace [tex]\( a \)[/tex] in the original integral form of [tex]\( f(x) \)[/tex] to get [tex]\( f(x) \)[/tex] in terms of [tex]\( b \)[/tex] and [tex]\( C \)[/tex]:
[tex]\[
f(x) = \frac{4 - b}{27} x^3 + b x + C
\][/tex]
Step 4: Use the y-intercept of [tex]\( C \)[/tex].
We know the y-intercept is -5, meaning [tex]\( f(0) = -5 \)[/tex]:
[tex]\[
f(0) = -5 \implies \frac{4 - b}{27} (0)^3 + b(0) + C = -5 \implies C = -5
\][/tex]
Step 5: Use the point (3, 5) to find [tex]\( b \)[/tex].
Substitute [tex]\( C = -5 \)[/tex] back into the equation and use the point [tex]\( (3, 5) \)[/tex]:
[tex]\[
f(3) = 5 = \frac{4 - b}{27} (3)^3 + b(3) - 5
\][/tex]
Simplify and solve for [tex]\( b \)[/tex]:
[tex]\[
5 = \frac{4 - b}{27} \cdot 27 + 3b - 5
\][/tex]
[tex]\[
5 = 4 - b + 3b - 5
\][/tex]
[tex]\[
5 = 4 + 2b - 5
\][/tex]
[tex]\[
5 = -1 + 2b
\][/tex]
[tex]\[
6 = 2b \implies b = 3
\][/tex]
Step 6: Find [tex]\( a \)[/tex].
Substitute [tex]\( b = 3 \)[/tex] back into the equation for [tex]\( a \)[/tex]:
[tex]\[
a = \frac{4 - 3}{9} = \frac{1}{9}
\][/tex]
Step 7: Write the final expression for [tex]\( f(x) \)[/tex].
Substitute [tex]\( a = \frac{1}{9} \)[/tex] and [tex]\( b = 3 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = \frac{1}{27} x^3 + 3x - 5
\][/tex]
Therefore, the equation of the curve [tex]\( C \)[/tex] is:
[tex]\[
f(x) = \frac{1}{27} x^3 + 3 x - 5
\][/tex]
