Can the mean value theorem be applied to this function: f(x)= ?2-x on the interval [-7,2].If this can be applied here find the points at which the slope of a line tangent to the graph is equal to the slope of the secant line at the endpoints of the given interval. Need some help with this one, anyone able to help me out I would appreciate it? I am somewhat confused with this problem. (?)=square root
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Mean Value Theorem
- Thread starter jon12
- Start date
f(x)= ?2-x on the interval [-7,2]. I will assume that you meant to put the whole expression 2-x under the square root.
This function is continuous on [-7,2] and differentiable on (-7,2) so that the MVT can be applied.
The slope of the secant line is
\(\displaystyle \frac{f(2)-f(-7)}{2+7}=-\frac{3}{9}=-\frac{1}{3}\).
The slope of the tangent line is
\(\displaystyle f'(c)=-\frac{1}{2\sqrt{2-c}}\)
Now set these two equal to each other, cross multiply and square both sides. The resulting equation is easy to solve and you should get c=-1/4.
Let me know if you need more details.
This function is continuous on [-7,2] and differentiable on (-7,2) so that the MVT can be applied.
The slope of the secant line is
\(\displaystyle \frac{f(2)-f(-7)}{2+7}=-\frac{3}{9}=-\frac{1}{3}\).
The slope of the tangent line is
\(\displaystyle f'(c)=-\frac{1}{2\sqrt{2-c}}\)
Now set these two equal to each other, cross multiply and square both sides. The resulting equation is easy to solve and you should get c=-1/4.
Let me know if you need more details.
D
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jon12 said:Can the mean value theorem be applied to this function: f(x)= ?2-x on the interval [-7,2].If this can be applied here find the points at which the slope of a line tangent to the graph is equal to the slope of the secant line at the endpoints of the given interval. Need some help with this one, anyone able to help me out I would appreciate it? I am somewhat confused with this problem. (?)=square root
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