Answer:
Global maxima are (3, 0) and (-3, 0),
Local minima is (0, -1.62)
Step-by-step explanation:
Here, the given function,
[tex]f(x) = -0.02x^4+0.36x^2-1.62[/tex]
Differentiating with respect to x,
[tex]f'(x) = -0.08x^3+0.72x[/tex]
For maxima or minima,
[tex]f'(x)=0[/tex]
[tex]\implies-0.08x^3+0.72x=0[/tex]
[tex]-0.08x(x^2-9)=0[/tex]
[tex]\implies x=0\text{ or }x=\pm 3[/tex]
Thus, the critical points of the function f(x) are 0, -3 and 3,
Since, f'(x) > 0 on the left side of x = -3 and f'(x) < 0 on the right side of x = -3,
⇒ x = -3 is local maxima,
Also, f(-3) = 0,
⇒ f(x) has maxima at (-3, 0),
f'(x) < 0 on the left side of x = 0 and f'(x) > 0 on the right side of x = 0,
⇒ x = 0 is the local minima,
Also, f(0) = -1.62
⇒ function f(x) has minima at (0, -1.62),
Hence, the global maxima are (3, 0) and (-3, 0),
Local minima is (0, -1.62).
f'(x) > 0 on the left side of x = 3 and f'(x) < 0 on the right side of x = 3,
⇒ x = 3 is local maxima,
Also, f(3) = 0,
⇒ function f(x) has maxima at (3, 0).
Note : function f(x) has no global minima because its end behaviour is,
[tex]\text{As }x\rightarrow \infty ; f(x)\rightarrow -\infty[/tex]
[tex]\text{As }x\rightarrow -\infty ; f(x)\rightarrow -\infty[/tex]
