Learn extreme value & fermat's theorem calculus, practice to find the critical points & numbers of the function and closed interval method.
A \(f\) has absolute maximum value at \(x=3\).
B \(f\) has absolute maximum value at \(x=1\).
C \(f\) has absolute minimum value at \(x=3\).
D \(f\) has absolute maximum value at \(x=5\).
A 3 local maxima and 2 local minima
B 2 local maxima and 2 local minima
C No local maxima and 3 local minima
D 3 local maxima and no local minima
A \(x=1,\;2,\,\dfrac{9}{2}\)
B \(x=0,\,5,\,13\)
C \(x=-2,\,-1,\,0\)
D \(x=0,\,1,\,\dfrac{7}{6}\)
If \(f\) has a local maximum or minimum at \('c'\) then \('c'\) must be a critical point
The collection of values in steps (1) and (2) will give the list of critical points.
A \(x=2,\,5\)
B \(x=1,\,3\)
C \(x=1,\,8\)
D \(x=5,\,-7\)
To find the absolute maximum and absolute minimum value of a function in closed interval [a, b]:
Some situations which can occur are shown graphically.
A Absolute maximum value=5, Absolute minimum value =10
B Absolute maximum value=2, Absolute minimum value =–10
C Absolute maximum value=1, Absolute minimum value =1
D Absolute maximum value= –7, Absolute minimum value =10
Then \(f'(x)\)\(=0\)
\(\Rightarrow\,3ax^2+2bx+c=0\)
Now, this equation can have two distinct roots or one root or no real root at all.
A 2 critical points
B 1 critical point
C No critical point
D 3 critical points