Q. Find a point c satisfying the conclusion of the MVT for the given function and interval.y(x)=x,[49,289]
Introduction: The Mean Value Theorem (MVT) states that if a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists at least one number c in the interval (a,b) such that f′(c)=b−af(b)−f(a). We need to apply this theorem to the function y(x)=x on the interval [49,289].
Verification: First, we need to verify that the function y(x)=x is continuous on [49,289] and differentiable on (49,289). The square root function is continuous and differentiable on any interval of positive numbers, so it satisfies the conditions of the MVT on our interval.
Average Rate of Change: Next, we calculate the average rate of change of the function over the interval [49,289]. This is given by (f(b)−f(a))/(b−a), where f(x)=x, a=49, and b=289.f(289)=289=17 (since 172=289)f(49)=49=7 (since 72=49)So, (f(b)−f(a))/(b−a)=(17−7)/(289−49)=10/240=1/24.
Derivative Calculation: Now we need to find the derivative of y(x)=x, which is y′(x)=2x1. We will set this equal to the average rate of change we just calculated.2c1=241
Finding Value of c: To find the value of c, we solve the equation 2c1=241 for c. 2c=24 c=12 c=122 c=144
Validity Check: We need to check that our found value of c is indeed in the open interval (49,289). Since 144 is between 49 and 289, it satisfies the condition.