If f(x) is differentiable for the closed interval [−1, 4] such that f(−1) = −3 and f(4) = 12, then there exists a value c, −1< c < 4 such that
a)f'(c)=3
b)f'(c)=0
C)f(c)=-15
d)f(c)=3
I understand that you are supposed to use the mean value theorem, but i dont understand how. If you could walk me through it step-by-step, that would be greatly appreciated.
if you google MVT you will see diagrams of what it means. All it amounts to is that if the conditions are met, there will be one or more places in the interval where the tangent line is parallel to the line joining the points at the ends of the interval.
In this case, the line joining the points on the graph at the ends of the interval has slope
(f(4)-f(-1))/(4-(-1))
= (12-(-3))/(4-(-1))
= 15/5
= 3
So, somewhere in [-1,4], the tangent line will have a slope of 3.
That is choice (a)
the function climbs uphill 15 from -3 to +12
in its jaunt of +5 from -1 to +4
that means that somewhere in between there the slope of the function is
slope = 15/5 = 3
so at some x = c between x = -1 and x=+4slope = f'(c) = 3
so
a)
To solve this question, we can use the Mean Value Theorem. The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a value c in (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
In our case, we are given that f(x) is differentiable for the closed interval [-1, 4] and f(-1) = -3 and f(4) = 12. We want to find a value c such that f'(c) equals a given value.
a) To find a value c such that f'(c) = 3, we need to find two points a and b such that the difference in f(x) values is 3 times the difference in x values. Let's set a = -1 and b = 4, which satisfies the given conditions. Applying the Mean Value Theorem, we have:
f'(c) = (f(4) - f(-1)) / (4 - (-1))
= (12 - (-3)) / (4 + 1)
= 15 / 5
= 3
Therefore, we have found a value c such that f'(c) = 3.
b) To find a value c such that f'(c) = 0, we need to find points a and b such that the difference in f(x) values is 0 (i.e., f(b) - f(a) = 0). This means that f(a) = f(b), which indicates that the function f(x) is constant on the interval [a, b]. Since f(x) is not a constant function (as f(-1) = -3 and f(4) = 12), there is no value of c such that f'(c) = 0.
Therefore, there is no value c such that f'(c) = 0.
c) To find a value c such that f(c) = -15, we can't directly apply the Mean Value Theorem since it deals with the derivative of the function. We need to use the Intermediate Value Theorem. The Intermediate Value Theorem states that if f(x) is continuous on the closed interval [a, b], then for every value y between f(a) and f(b), there exists a value c in [a, b] such that f(c) = y.
In our case, f(x) is continuous on [-1, 4]. We are given that f(-1) = -3 and f(4) = 12, and we want to find a value c such that f(c) = -15. Since -15 is between -3 and 12, there exists a value c in the interval [-1, 4] such that f(c) = -15.
Therefore, there exists a value c, -1 < c < 4, such that f(c) = -15.
d) To find a value c such that f(c) = 3, we can once again use the Intermediate Value Theorem. Since we are given that f(-1) = -3 and f(4) = 12, we can see that f(4) > 3 and f(-1) < 3. Therefore, by the Intermediate Value Theorem, there exists a value c in the interval [-1, 4] such that f(c) = 3.
Therefore, there exists a value c, -1 < c < 4, such that f(c) = 3.
In conclusion:
a) f'(c) = 3 (True)
b) f'(c) = 0 (False)
c) f(c) = -15 (True)
d) f(c) = 3 (True)
To apply the Mean Value Theorem to this problem, we need to verify that the given function meets the conditions required by the theorem. The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a value c in the interval (a, b) such that f'(c) = (f(b) - f(a))/(b-a).
In this case, we are given that f(x) is differentiable for the closed interval [-1, 4]. So, the function meets this condition.
Now, let's see if we can find a value c in the interval (-1, 4) that satisfies each of the given options:
a) To find a value c such that f'(c) = 3, we're looking for a point on the graph where the slope of the tangent line equals 3. We can use the Mean Value Theorem to find this point. First, let's find the values of f(-1) and f(4): f(-1) = -3 and f(4) = 12. Now, let's substitute these values into the formula:
f'(c) = (f(4) - f(-1))/(4 - (-1))
f'(c) = (12 - (-3))/(4 + 1)
f'(c) = 15/5
f'(c) = 3
So, for option a), we have found a value c such that f'(c) = 3.
b) For f'(c) = 0, we're looking for a point where the slope of the tangent line is 0. Similarly, using the same process as above, we find:
f'(c) = (f(4) - f(-1))/(4 - (-1))
f'(c) = (12 - (-3))/(4 + 1)
f'(c) = 15/5
f'(c) = 3
Since f'(c) ≠ 0, we cannot find a value c for option b).
c) To check if there exists a value c such that f(c) = -15, we need to evaluate f(c) at different c values. However, the given options do not suggest specific c values, so we cannot determine if f(c) = -15 based on the Mean Value Theorem alone.
d) Similarly, we don't have enough information to determine if f(c) = 3 based on the given options.
In conclusion, based on the information given, we have only found a value c such that f'(c) = 3 from the Mean Value Theorem. The theorem does not provide direct information about the values of f(c) for a given c.