Which of the following is a true statement?
Question 1 ?Answer
a.
The function h(x)=-x^2
has an absolute minimum at x=0
.
b.
The function l(x)=|x+2|
is differentiable at x=-2
. ?
c.
The function q(x)=x^4
is decreasing for all values of x
.
d.
The function f(x)=-x^3
has no critical points.
e.
The function p(x)=sin(x)
has infinitely many critical points.
f.
The function g(x)=-cos(x)
has no critical points.
The Correct Answer and Explanation is :
The correct statement is:
e. The function p(x) = sin(x) has infinitely many critical points.
Explanation:
Let’s analyze each option in detail:
a. The function h(x) = -x² has an absolute minimum at x = 0.
This is false. The function ( h(x) = -x^2 ) is a downward-opening parabola, which means it has an absolute maximum at ( x = 0 ), not a minimum. As ( x ) moves away from zero, the function values become increasingly negative, so no minimum exists at ( x = 0 ).
b. The function l(x) = |x + 2| is differentiable at x = -2.
This is false. The function ( l(x) = |x + 2| ) has a “corner” at ( x = -2 ), which means it is not differentiable there. The derivative from the left does not equal the derivative from the right at this point, leading to a discontinuity in the slope.
c. The function q(x) = x⁴ is decreasing for all values of x.
This is false. The function ( q(x) = x^4 ) is increasing for ( x > 0 ) and decreasing for ( x < 0 ). At ( x = 0 ), the function is flat, but it is not globally decreasing. The behavior depends on the sign of ( x ).
d. The function f(x) = -x³ has no critical points.
This is false. The function ( f(x) = -x³ ) does have a critical point at ( x = 0 ). A critical point occurs when the derivative is zero or undefined. The derivative of ( f(x) = -x³ ) is ( f'(x) = -3x² ), which equals zero at ( x = 0 ), so this is a critical point.
e. The function p(x) = sin(x) has infinitely many critical points.
This is true. The function ( p(x) = \sin(x) ) has critical points where the derivative is zero. The derivative of ( \sin(x) ) is ( \cos(x) ), and ( \cos(x) = 0 ) at ( x = \frac{\pi}{2} + n\pi ), where ( n ) is any integer. Hence, there are infinitely many critical points of the form ( x = \frac{\pi}{2} + n\pi ).
f. The function g(x) = -cos(x) has no critical points.
This is false. The derivative of ( g(x) = -\cos(x) ) is ( g'(x) = \sin(x) ). ( g'(x) = 0 ) at ( x = n\pi ), where ( n ) is an integer, so there are critical points at these locations.
Conclusion:
The correct answer is e.