To determine the nature of the function
f(x)=x3−3x2+4x on real numbers, we need to analyze its first derivative. The first derivative of a function provides information about its increasing or decreasing behavior.
Let's find the first derivative of
f(x) :
f′(x)=‌(x3−3x2+4x) By applying the power rule of differentiation:
f′(x)=3x2−6x+4Next, we need to analyze the sign of
f′(x). The critical points of the function occur where
f′(x)=0.
Solving for critical points:
3x2−6x+4=0 We can solve this quadratic equation using the quadratic formula:
x=‌ where
a=3,b=−6, and
c=4.Substituting these values into the formula gives us:
‌x=‌‌x=‌‌x=‌‌x=1±‌ The solutions are complex, which means there are no real critical points. Therefore, there are no changes in the sign of
f′(x) over the real numbers.
Now, let's analyze the sign of
f′(x)=3x2−6x+4 :
The quadratic expression
3x2−6x+4 is always positive for all real numbers because its discriminant
% (which is the value under the square root in the quadratic formula ) is negative
(∆=b2−4ac=36−48=−12). Since
f′(x) is always positive and there are no real roots for the equation, the function is always increasing.
Therefore, the nature of the function
f(x)=x3−3x2+4x on the real numbers is strictly increasing.
The correct answer is:
Option C