For which value? ### Problem StatementFor which \( x \) value does \( f(x) = x^3 - 3x^2 + 1 \) have a local maximum?### Options- \( \circ \) 0- \( \circ \) -1- \( \circ \) 1- \( \circ \) -2- \( \circ \) 2To determine the \( x \) value where the function \( f(x) = x^3 - 3x^2 + 1 \) has a local maximum, we need to find the critical points by setting the derivative \( f'(x) \) equal to zero and analyzing the second derivative to determine the nature of these points.

Name: For which value? ### Problem StatementFor which \( x \) value does \(
Uploaded: 2024-03-20T07:06:16.000Z
Channel: Bartleby
Description: For which value? ### Problem StatementFor which \( x \) value does \( f(x) = x^3 - 3x^2 + 1 \) have a local maximum?### Options- \( \circ \) 0- \( \circ \) -1- \( \circ \) 1- \( \circ \) -2- \( \circ \) 2To determine the \( x \) value where the funct

Given: The function fx=x3-3x2+1, To find: point of Local maximumExplanation: It is known that A function f at a point x=a has Local maximum if f'a=0 and f''a<0 local minimum if f'a=0 and f''a>0 Here the function is fx=x3-3x2+1 Differentiating f with respect to x, f'x=ddxx3-3x2+1⇒f'x=3x2-6x Solving f'x=0 for x, f'x=0⇒3x2-6x=0⇒3xx-2=0⇒x=0,2 Now again differentiating f'x with respect to x f''x=ddx3x2-6x⇒f''x=6x-6 Finding the value of f''0, f''2, f''0=6×0-6=-6<0 f''2=6×2-6=12-6=6>0. Hence we have for x=0, f'x=0 and f''x<0 Therefore x=0 is a point of local maximum. Answer: 0 is correct option.