Find all critical points of the function f(x)=7x2−2x+23x2.(Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. If the function does not have any critical points, enter DNE.)critical points:
Q. Find all critical points of the function f(x)=7x2−2x+23x2.(Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. If the function does not have any critical points, enter DNE.)critical points:
Calculate Derivative: To find the critical points of the function f(x)=7x2−2x+23x2, we need to find the values of x where the derivative f′(x) is zero or undefined.
Apply Quotient Rule: First, we calculate the derivative of f(x) using the quotient rule, which states that if f(x)=h(x)g(x), then f′(x)=(h(x))2g′(x)h(x)−g(x)h′(x). Here, g(x)=3x2 and h(x)=7x2−2x+2.
Simplify Numerator: Compute the derivative of g(x)=3x2, which is g′(x)=6x.
Find Critical Points: Compute the derivative of h(x)=7x2−2x+2, which is h′(x)=14x−2.
Factor Common Term: Apply the quotient rule to find f′(x):f′(x)=(7x2−2x+2)2(6x⋅(7x2−2x+2)−3x2⋅(14x−2)).
Solve for x: Simplify the numerator of f′(x):Numerator=6x⋅(7x2−2x+2)−3x2⋅(14x−2)=42x3−12x2+12x−42x3+6x2=−12x2+12x+6x2=−6x2+12x.
Check Denominator: Set the simplified numerator equal to zero to find the critical points:−6x2+12x=0.
Identify Critical Points: Factor out the common term:−6x(x−2)=0.
Identify Critical Points: Factor out the common term:−6x(x−2)=0.Set each factor equal to zero and solve for x:−6x=0 gives x=0.x−2=0 gives x=2.
Identify Critical Points: Factor out the common term:−6x(x−2)=0. Set each factor equal to zero and solve for x:−6x=0 gives x=0.x−2=0 gives x=2. Check if the denominator of f′(x) becomes zero at x=0 or x=2, which would make the derivative undefined:(7x2−2x+2)2 at x=0 is x1, which is not zero.(7x2−2x+2)2 at x=2 is x4, which is x5, which is not zero.
Identify Critical Points: Factor out the common term:−6x(x−2)=0. Set each factor equal to zero and solve for x:−6x=0 gives x=0.x−2=0 gives x=2. Check if the denominator of f′(x) becomes zero at x=0 or x=2, which would make the derivative undefined:(7x2−2x+2)2 at x=0 is x1, which is not zero.(7x2−2x+2)2 at x=2 is x4, which is x5, which is not zero. Since the denominator of f′(x) does not become zero at x=0 or x=2, the critical points are x=0 and x=2.