Application of Derivatives Test - 1

Explanation:

Since the graph cuts the lines y = -1 and y = 1 , therefore ,it must cut y = 0 atleast once as the graph is a continuous curve in this case.

Explanation:

f ‘ (x) = 0 ⇒ f (x) is constant in ( 0 , 1 ) and also in ( 2, 4 ). But this does not mean that f ( x) has the same value in both the intervals . However , if f ( c ) =f ( d ) , where c ∈( 0 , 1 ) and d ∈ ( 2, 4) then f ( x ) assumes the same value at all x ∈( 0 ,1 ) U (2, 4 ) and hence f is a constant function.

Explanation:

In case of a strict decreasing function , slope of tangent and hence derivative is either negative or zero because decreasing function change sign from positive to negative and making obtuse angle measured clockwise.Hence negative and will be zero at peak.

Explanation:

f (x) = 2 – 3x ⇒ f ‘ (x) = - 3 < 0 for all x ∈ R . so, f is strictly decreasing function.

Explanation:

 f(x) = x²- 2x

⇒f'(x) = 2x - 2 = 2(x - 1) 

So , f ( x) is increasing if 2 (x-1) ≥ 0 , i.e.if x ≥ 1

Explanation:

f(x) = x²
⇒ f'(x) = 2x for all x in R.

Since f ‘(x) = 2x > 0 for x >0, and f ‘ (x) = 2x < 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing . Infact , f is strict increasing on [ 0 , ∞ ) and strict decreasing on (- ∞,0 ].

Explanation:

f ( x) = mx + c is strict decreasing on R

if f ‘ ( x) < 0 i.e. if m < 0 .

Explanation:

Here ,f′(x)=3x²−12x+9

= 3 (x−1) (x−3) < 0 ⇔ x ∈ (1,3)