find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)
f(x)=√(4-x), [-5,4]
![find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)f(x)=√(4-x), [-5,4] class=](https://us-static.z-dn.net/files/de5/e6ce7fb94d04a5b6d3815028f7a8c742.jpg)
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Work Shown:
Compute the function value at the endpoints
[tex]f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\[/tex]
With a = -5 and b = 4, we have
[tex]f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\[/tex]
So,
[tex]f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\[/tex]
Use algebra to solve for c
[tex]-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\[/tex]