Video: Finding the Values That Satisfy the Mean Value Theorem for a Radical Function in a Given Interval

For the function 𝑓(π‘₯) = √π‘₯, find all the possible values of 𝑐 that satisfy the mean value theorem over the interval [4, 9].

03:46

Video Transcript

For the function 𝑓 of π‘₯ is equal to the square root of π‘₯, find all the possible values of 𝑐 that satisfy the mean value theorem over the closed interval from four to nine.

The question gives us a function. 𝑓 of π‘₯ is equal to the square root of π‘₯. And it wants us to find all possible values of 𝑐 which satisfy the mean value theorem over the closed interval from four to nine. We recall that the mean value theorem tells us that if a function 𝑓 of π‘₯ is continuous on the closed interval from π‘Ž to 𝑏 and it’s differentiable on the open interval from π‘Ž to 𝑏. Then, there must exist a 𝑐 in the open interval from π‘Ž to 𝑏 such that the derivative function 𝑓 prime evaluated at 𝑐 is equal to 𝑓 evaluated at 𝑏 minus 𝑓 evaluated at π‘Ž all divided by 𝑏 minus π‘Ž.

The question wants us to find all the possible values of 𝑐 which would work when we apply the mean value theorem to our function 𝑓 of π‘₯ is equal to the square root of π‘₯ on the closed interval from four to nine. So, we’ll set 𝑓 of π‘₯ equal to the square root of π‘₯, π‘Ž equal to four, and 𝑏 equal to nine. This means we’re looking for all possible values of 𝑐 in the open interval from π‘Ž to 𝑏 such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž. First, we notice that 𝑐 being in the open interval from π‘Ž to 𝑏, when π‘Ž is four and 𝑏 is nine, is the same as saying that four is less than 𝑐 is less than nine.

Next, we want to calculate our derivative function, 𝑓 prime of π‘₯. We have our derivative function 𝑓 prime of π‘₯ is equal to the derivative of the square root of π‘₯ with respect to π‘₯. And we can rewrite this as π‘₯ to the power of a half. We then recall, for constants π‘Ž and 𝑛, to differentiate π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯, we get π‘Žπ‘› multiplied by π‘₯ to the power of 𝑛 minus one. We multiply by the exponent and then reduce the exponent by one. So, differentiating π‘₯ to the power of a half, we multiplied by a half and reduced the exponent by one. This gives us a half multiplied by π‘₯ to the power of a half minus one, which simplifies to give us a half π‘₯ to the power of negative a half.

We’re now ready to try and find all the possible values of 𝑐 which satisfy the mean value theorem on the closed interval from four to nine for 𝑓 of π‘₯ is equal to the square root of π‘₯. We want to find the values of 𝑐 between four and nine such that 𝑓 prime of 𝑐 is equal to 𝑓 of nine minus 𝑓 of four divided by nine minus four. We know that 𝑓 prime of 𝑐 is equal to a half 𝑐 to the power of negative a half. And we know that 𝑓 of π‘₯ is equal to the square root of π‘₯. Therefore, 𝑓 of nine minus 𝑓 of four over nine minus four is the square root of nine minus the square root of four divided by nine minus four. We have the square root of nine is three, and the square root of four is two. So, this simplifies to three minus two over five, which is equal to one-fifth.

So, we want to find all the values of 𝑐 such that a half 𝑐 to the power of negative a half is equal to one-fifth and 𝑐 is between four and nine. So, let’s solve this equation for 𝑐. We multiply both sides of the equation by two to get 𝑐 to the power of negative a half is equal to two-fifths. Next, we take the reciprocal of both sides of our equation to get that 𝑐 to the power of a half is equal to five over two. Finally, we square both sides of our equation to get that 𝑐 is equal to five over two all squared, which is equal to 25 divided by four.

And just to be safe, we check that this value is inside our open interval. We see that its decimal expansion is 6.25, which is between four and nine. Therefore, we’ve shown, for the function 𝑓 of π‘₯ is equal to the square root of π‘₯, all the possible values of 𝑐 which satisfy the mean value theorem on the closed interval from four to nine is just when 𝑐 is equal to 25 divided by four.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.