Finding Critical Point Functions
find all of the critical points and local maximums and minimums of each function
f(x)=〖2x〗^2-12x+7
find all critical points and local extremes of each function on the given intervals.
f(x)=X^2-3x+5 on the entire real number line.
find all critical points and local extremes of each function on the given intervals.
f(x)=□(1/(x^2+1)) on the entire real number line.
verify that the hypotheses of Rolle's Theorem are satisfied for each of the functions on the given intervals, and find the value of the number(s) "c" that Rolle's Theorem promises.
F(x) =x^2 on [-2, 2]
verify that the hypotheses of the Mean Value Theorem are satisfied for each of the functions on the given intervals, and find the number(s) "c" that the Mean Value Theorem guarantees.
f(x)=sin(x) on [0, π/2]
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1. find all of the critical points and local maximums and minimums of each function
Solution:
To find critical points, put f'(x) = 0
4x-12 = 0
4x = 12
x = 3
Critical point is x = 3
Find f''(x)
Since f''(3) > 0 so function has local minimum at x = 3
Local minimum is at (3, -11).
2. Find all critical points and local extremes of each function on the given intervals.
on the entire real number line.
Solution:
To find critical points, put f'(x) = 0
= 0
2x = 3
x = 3/2
Critical point is x = 3/2
Find f''(x)
Since f''(3/2) > 0 so function has local minimum at x = 3/2
Local minimum is at ( , ).
3. find all critical points and local extremes of each function on the given intervals.
on the entire real number line.
Solution:
To find critical points, put f'(x) = 0
= 0
This gives x = 0
Critical point is x = 0
Find f''(x)
Now find the value of f"(0)
Since f''(0) < 0 so function has local maximum at x = 0
Local maximum is at (0, 1).
4. Verify that the hypotheses of Rolle's Theorem are satisfied for each of the functions on the given intervals, and find the value of the number(s) "c" that Rolle's Theorem promises.
F(x) = on [-2, 2]
Solution:
F(x) is continuous over [-2, 2] since polynomials are continuous over their domains, F(x) is differentiable over (-2, 2) since polynomials are differentiable over the real numbers.
F(-2) = (-2)2 = 4
F(2) = (2)2 = 4
And F(-2) = F(2)
Thus Rolle's Theorem applies.
F'(x) = 2x
F'(c) = 2c
Put F'(c) = 0
2c = 0
c = 0
Answer: c = 0
5. Verify that the hypotheses of the Mean Value Theorem are satisfied for each of the functions on the given intervals, and find the number(s) "c" that the Mean Value Theorem guarantees.
on [0,
Solution:
sinx is continuous in [0, and differentiable in the domain (0, π/2).
Now,
Put f'(c) =
cos(c) =
c = 50.5°
Answer: c = 50.5°
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