Question

Find the interval in which the function f(x)=3/2x4-4x3-45x2+51 is (a) strictly increasing (b)strictly decreasing.

Answer 1

The Function f(x) is

f(x)= $\frac{3}{2}x^4-4 x^3-45 x^2+51$

To find the intervals in which the function is increasing or Decreasing , we have to differentiate the function first

f'(x)= $\frac{3}{2} \times 4 x^3 - 4 \times 3 x^2 -45 \times 2 x$

f'(x)= $6 x^3 - 12 x^2 -90 x$

Now, put f'(x)=0

$6 x^3 - 12 x^2 -90 x$=0

Dividing both sides by 6 x, we get

x=0

and

$x^2 - 2 x -15$=0

Factorizing that is Splitting the middle term, we get

$x^2 - 5 x + 3 x -15$=0

$x(x-5)+3(x -5)$=0

(x+3)(x-5)=0

Gives , x= -3 and x=5 and x=0

x= 0,-3,5→→Critical points

Draw the number line and mark -3,0 and 5 on it . And check where the function is increasing or decreasing substitute real numbers between these points

→1 lies between 0 and 5, as you will see that ,f'(1)<0.So in this interval f(x) is decreasing.

Similarly ,you can check and find the intervals where the function is Strictly increasing and Strictly decreasing.

Strictly increasing =(-3,0) ∪ (5, ∞]

Strictly Decreasing = [-∞,-3)∪(0, 5)

Answer 2

Answer:

Step-by-step explanation: