Given the function # f(x) = 8 sqrt{ x} + 1#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,10] and find the c?

Answer 1

Please see below.

The Mean Value Theorem has two hypotheses:

H1 : #f# is continuous on the closed interval #[a,b]#
H2 : #f# is differentiable on the open interval #(a,b)#.

A hypothesis is satisfied if it is true.

So we need to determine whether

#f(x) = 8sqrtx+1# is continuous on the closed interval #[1,10]#. It is.
(#f# is continuous on its domain, #[0,oo)#, so it is continuous on #[1,10]#.)

and whether

#f(x) = 8sqrtx+1# is differentiable on the open interval #(1,10)#. It is.
(Remember that "differentiable" means the derivative exists.) #f'(x) = 4/sqrtx# exists for all #x > 0#. So, #f# is differentiable on #(1,10)#.)

A very common way of stating the Mean Value Theorem gives the conclusion as:

There is a #c# in #(a,b)# such that #f'(c) = (f(b)-f(a))/(b-a)#
A popular exercise (calculus and algebra) is to ask students to find the value or values of #c# that satisfy this conclusion.
To do this, find #f'(x)#, do the arithmetic to find #(f(b)-f(a))/(b-a)# and solve the equation
#f'(x) = (f(b)-f(a))/(b-a)# on the interval (a,b)#
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Answer 2

To determine if the function ( f(x) = 8\sqrt{x} + 1 ) satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval ([1,10]), first, ensure that ( f(x) ) is continuous on ([1,10]) and differentiable on ( (1,10) ). Since the square root function is continuous and differentiable for all ( x \geq 0 ), ( f(x) ) satisfies these conditions on the given interval.

Next, compute the derivative of ( f(x) ) to find ( f'(x) ). The derivative of ( 8\sqrt{x} + 1 ) with respect to ( x ) is ( \frac{4}{\sqrt{x}} ).

Now, evaluate ( f'(x) ) at the endpoints of the interval, ( x = 1 ) and ( x = 10 ).

( f'(1) = \frac{4}{\sqrt{1}} = 4 ) and ( f'(10) = \frac{4}{\sqrt{10}} ).

Since ( f'(1) ) and ( f'(10) ) are both finite, ( f'(x) ) is continuous on ([1,10]).

By the Mean Value Theorem, there exists a number ( c ) in the open interval ((1,10)) such that ( f'(c) ) is equal to the average rate of change of ( f(x) ) over the interval ([1,10]).

To find ( c ), set ( f'(c) ) equal to the average rate of change of ( f(x) ) over ([1,10]), which is given by:

[ \frac{f(10) - f(1)}{10 - 1} ]

[ = \frac{8\sqrt{10} + 1 - (8\sqrt{1} + 1)}{9} ]

[ = \frac{8\sqrt{10} - 8}{9} ]

[ = \frac{8}{9}(\sqrt{10} - 1) ]

Setting ( f'(c) ) equal to this value:

[ \frac{4}{\sqrt{c}} = \frac{8}{9}(\sqrt{10} - 1) ]

Solve for ( c ) to find the point where the tangent line is parallel to the secant line on ([1,10]).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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