Question

\[\text{Let } f(x) =x^3 -\frac{3}{2}x^2-6x+5\]\[\text{Prove that}\]\[2 \sqrt{3} < \int\limits_{-2}^{0} \sqrt{f(x)}\ dx < \sqrt{34}\]

Answer 1

Hi Davidc

f(x) is a continuous function on the given interval [-2,0]
all u need is to find max and min of f(x) on the interval [-2,0] and for doing that u must Compute f(x) at the critical point(s) and at the boundaries of the closed interval.

boundaries : f(-2)=3 and f(0)=5

critical point(s) : f'(x)=0 ----> x^2-x-2=0 ----> x=-1 and x=2
and x=-1 is on our interval ---------> f(-1)=17/2

so u have
max of f(x) ----->f(-1)=17/2
min of f(x) ----->f(-2)=3
and this means
\[3 \le f(x) \le \frac{17}{2} \ \ \ \rightarrow \ \ \ \sqrt{3} \le \sqrt{f(x)} \le \sqrt{\frac{17}{2}} \]

now what will u get?

Answer 2

Thank you :)\[2 \sqrt{3} < \int\limits_{-2}^{0} \sqrt{f(x)}\ dx < \sqrt{34}\]