Test: Calculus- 2

We need absolute maximum of 
f(x) = 2x² - 2x + 6 
in the interval [0, 2]
First find local maximum if any by putting f'{x) = 0

is a point of local minimum. So there is no point of local maximum.
Now tabulate the values of f at end point of interval and at local maximum if any (in this case no point of local maximum).

Clearly the absolute maxima is at x = 2 and absolute maximum value is 10.

Comparing with area under the standared normal curve from -∝ to 0.
We get

So, the required integral

For the function 25/8x the minimum value will come when x is maximum since it is a decreasing function.
The maximum value of x is 3/2.
 the function has the value 


But since for this function  this function we get the minimum of this function which is 
Now comparing the minimum value 2.0833 of the first function with minimum value 2.166 of the second function, we get the overall minimum of this function to be 2.0833   which is option (b).

Now
x = 0

∴ f(x) has a minimum at x = 0

So x = 2 is a saddle point (point of inflection)
∴ f(x) has no extremum at x = 2. So f(x) has only one point of extremum (at x = 0).

⇒ y = 1

∴ There is only one maxima and only two minima for this function.

So at x = 1 we have a relative maxima.

We need absolute maximum of 
f(x) = x³ - 9x² + 24x + 5 in the interval [1,6]
First find local maximum if any by putting f'(x) = 0. 

Now tabulate the values of f at end pt. of interval and at local maximum pt., to find absolute maximum in given range, as shown below:

Clearly the absolute m axim a is at x = 6 and absolute maximum value is 41.

Hence f(n) have maximum value at n = 1

Let,

As f(x) is square of hence its minimum value be 0 where at x = 1.

Hence f(x) has minimum value at x = 3 which is 28.

Intermediate value theorem states that if a function is continuous and f(a) • f(b) < 0, then surely there is a root in (a, b). The contrapositive of this theorem is that if a function is continuous and has no root in (a, b) then surely f(a) . f{b) ≥ 0. But since it is given that there is no root in the closed interval [a ,b] it means f(a) . f(b) ≠ 0.
So surely f(a) . f(b) > 0.