Using the critical point concept, it is found that a = -3 and b = 7.
[tex]f^{\prime}(x) = 0[/tex]
In this problem, the function is:
[tex]f(x) = x^3 + ax^2 + b[/tex]
Hence, the derivative is:
[tex]f^{\prime}(x) = 3x^2 + 2ax[/tex]
Then:
[tex]f^{\prime}(x) = 0[/tex]
[tex]3x^2 + 2ax = 0[/tex]
[tex]x(3x + 2a) = 0[/tex]
[tex]x = 0[/tex]
[tex]3x + 2a = 0[/tex]
[tex]3x = -2a[/tex]
[tex]x = -\frac{2a}{3}[/tex]
Since the critical point is at x = 2, we have that:
[tex]-\frac{2a}{3} = 2[/tex]
[tex]-2a = 6[/tex]
[tex]2a = -6[/tex]
[tex]a = -3[/tex]
Then:
[tex]f(x) = x^3 - 3x^2 + b[/tex]
Critical point at (2,3) means that when [tex]x = 2, y = 3[/tex], then:
[tex]3 = 2^3 - 3(2)^2 + b[/tex]
[tex]8 - 12 + b = 3[/tex]
[tex]b = 7[/tex]
You can learn more about the critical point concept at https://brainly.com/question/2256078