If the largest interval of x in which the function f(x)=x3-3x+1 is decreasing is (a , b) , then the value of a+2b is equal to

Question

If the largest interval of x in which the function f(x)=x3-3x+1 is decreasing is (a , b) , then the value of a+2b is equal to (A) -1 (B) 0 (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

As f'(x)=3x2-3=3(x - 1)(x + 1) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-vre8if8cpc37.png" /> i.e. the largest interval in which f(x) decreases is (- 1,1)⇒ a=-1 b=1 ∴ a+2b=1

Q. If the largest interval of $x$ in which the function $f (x) = x^{3} - 3 x + 1$ is decreasing is $(a, b)$ , then the value of $a + 2 b$ is equal to

$- 1$

$0$

$1$

$2$

As $f^{^{'}} (x) = 3 x^{2} - 3 = 3 (x - 1) (x + 1)$

i.e. the largest interval in which $f (x)$ decreases is $(- 1, 1) \Rightarrow a = - 1$
$b = 1$
$∴ a + 2 b = 1$

Q. If the largest interval of x in which the function f(x)=x3−3x+1 is decreasing is (a,b) , then the value of a+2b is equal to

−1

0

1

2

As f′(x)=3x2−3=3(x−1)(x+1) i.e. the largest interval in which f(x) decreases is (−1,1)⇒a=−1 b=1 ∴a+2b=1

Q. If the largest interval of $x$ in which the function $f (x) = x^{3} - 3 x + 1$ is decreasing is $(a, b)$ , then the value of $a + 2 b$ is equal to

$- 1$

$0$

$1$

$2$

As $f^{^{'}} (x) = 3 x^{2} - 3 = 3 (x - 1) (x + 1)$

i.e. the largest interval in which $f (x)$ decreases is $(- 1, 1) \Rightarrow a = - 1$
$b = 1$
$∴ a + 2 b = 1$