Question #d530f

1 Answer
Sep 20, 2015

See the explanation.

Explanation:

Lets write function in the form:

f(x)=4-x^2, x in[-2,2]
f(x)=-4+x^2, x in (-oo,-2)uu(+2,+oo)

1st derivative:

f'(x)=-2x, x in[-2,2]
f'(x)=2x, x in (-oo,-2)uu(+2,+oo)

f'(x)=0
-2x=0 <=> x=0, x in [-2,2], x is critical point
2x=0 <=> x=0, x !in (-oo,-2)uu(+2,+oo), x is not critical point

Sign of 1st derivative:
AAx in (-oo,-2) f'(x)=2x<0, f is decreasing
AAx in (-2,0) f'(x)=-2x>0, f is increasing
AAx in (0,+2) f'(x)=-2x<0, f is decreasing
AAx in (+2,+oo) f'(x)=2x>0, f is increasing

f'(x)=2x on the interval [-2,2] is continuous function, f'(x) changes sign in x=0 hence function f(x) has maximum value in x=0 and f_max=f(0)=4.

In points x=-2 and x=2 may exist discontinuities of the 1st order, so we have to check if the function f(x) is continuous in these points.

f(-2)=0
lim_(x->-2_-)f(x)=lim_(x->-2_-)(-4+x^2)=A
x=-2-epsilon
A=lim_(epsilon->0)(-4+(-2-epsilon)^2)=lim_(epsilon->0)(-4+4+4epsilon+epsilon^2)
=lim_(epsilon->0)(4epsilon+epsilon^2)=0

We prove that lim_(x->2_-)f(x)=f(-2) and hence function is continuous at the point x=-2.
f'(x) changes sign in x=-2 and function has minimum value f_min=f(-2)=0.

f(2)=0
lim_(x->2_+)f(x)=lim_(x->2_+)(-4+x^2)=B
x=2+epsilon
B=lim_(epsilon->0)(-4+(2+epsilon)^2)=lim_(epsilon->0)(-4+4+4epsilon+epsilon^2)
=lim_(epsilon->0)(4epsilon+epsilon^2)=0

Again, we prove that lim_(x->2_+)f(x)=f(2) and hence function is continuous at the point x=2.
f'(x) changes sign in x=2 and function has minimum value f_min=f(2)=0.