Answer:
Intercepts:
x = 0, y = 0
x = 1.77, y = 0
x = 2.51, y = 0
Critical points:
x = 1.25, y = 4
x = 2.17 , y = -4
x = 2.8, y = 4
Inflection points:
x = 0.81, y = 2.44
x = 1.81, y = -0.54
x = 2.52, y = 0.27
Step-by-step explanation:
We can find the intercept by setting f(x) = 0
[tex]4sin(x^2) = 0[/tex]
[tex]sin(x^2) = 0 [/tex]
[tex]x^2 = n\pi[/tex] where n = 0, 1, 2,3, 4, 5,...
[tex]x = \sqrt(n\pi)[/tex]
Since we are restricting x between 0 and 3 we can stop at n = 2
So the function f(x) intercepts at y = 0 and x:
x = 0
x = 1.77
x = 2.51
The critical points occur at the first derivative = 0
[tex]f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0[/tex]
[tex]xcos(x^2) = 0[/tex]
[tex]x = 0[/tex] or
[tex]cos(x^2) = 0[/tex]
[tex]x^2 = \frac{\pi}{2} + n\pi[/tex] where n = 0, 1, 2, 3
[tex]x = \sqrt{\pi(n+1/2)}[/tex]
Since we are restricting x between 0 and 3 we can stop at n = 2
So our critical points are at
x = 1.25, [tex]y = f(1.25) = 4sin(1.25^2) = 4[/tex]
x = 2.17 , [tex]y = f(2.17) = 4sin(2.17^2) = -4[/tex]
x = 2.8, [tex]y = f(2.8) = 4sin(2.8^2) = 4[/tex]
For the inflection point, we can take the 2nd derivative and set it to 0
[tex]f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0[/tex]
[tex]cos(x^2) = 2x^2sin(x^2)[/tex]
[tex]tan(x^2) = \frac{1}{2x^2}[/tex]
We can solve this numerically to get the inflection points are at
x = 0.81, [tex]y = f(0.81) = 4sin(0.81^2) = 2.44[/tex]
x = 1.81, [tex]y = f(1.81) = 4sin(1.81^2) = -0.54[/tex]
x = 2.52, [tex]y = f(2.52) = 4sin(2.52^2) = 0.27[/tex]
