Sketch ( y = |x^2 - 4| - 1 ). How many x-intercepts?
Stationary points occur when ( g'(x)=0 ). ( g(x) = 2f(1-x) + 1 ) ( g'(x) = 2 \cdot f'(1-x) \cdot (-1) = -2 f'(1-x) ) Set ( g'(x)=0 \implies f'(1-x)=0 ).
Now ( f'(x)=3x^2-3 = 3(x^2-1) ). So ( f'(1-x)=0 \implies (1-x)^2 - 1 =0 \implies (1-x)^2=1 ) ( \implies 1-x = \pm 1 \implies x=0 ) or ( x=2 ).
The graph of ( y = \cos x ) is transformed to ( y = 3\cos(2x - \pi) + 1 ). Describe the sequence.
Sketch ( y = |x^2 - 4| - 1 ). How many x-intercepts?
Stationary points occur when ( g'(x)=0 ). ( g(x) = 2f(1-x) + 1 ) ( g'(x) = 2 \cdot f'(1-x) \cdot (-1) = -2 f'(1-x) ) Set ( g'(x)=0 \implies f'(1-x)=0 ). transformation of graph dse exercise
Now ( f'(x)=3x^2-3 = 3(x^2-1) ). So ( f'(1-x)=0 \implies (1-x)^2 - 1 =0 \implies (1-x)^2=1 ) ( \implies 1-x = \pm 1 \implies x=0 ) or ( x=2 ). Sketch ( y = |x^2 - 4| - 1 )
The graph of ( y = \cos x ) is transformed to ( y = 3\cos(2x - \pi) + 1 ). Describe the sequence. transformation of graph dse exercise
