Suppose f is a continuous function and f prime(x) exists everywhere. If f(2)=10 and f prime(x) ≥-3 for all x, then the smallest possible value for f(4) is

Question

Suppose f is a continuous function and f prime(x) exists everywhere. If f(2)=10 and f prime(x) ≥-3 for all x, then the smallest possible value for f(4) is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

For minimum value of f (4) ( f (4)- f (2)/4-2)= min ( f prime( x )) ⇒ f (4)- f (2)=2 ⋅(-3) ⇒ f (4)=-6+10=4

Q. Suppose $f$ is a continuous function and $f^{'} (x)$ exists everywhere. If $f (2) = 10$ and $f^{'} (x) \geq - 3$ for all $x$ , then the smallest possible value for $f (4)$ is

1

2

3

4

For minimum value of $f (4)$
$\frac{f ( 4 ) - f ( 2 )}{4 - 2} = min (f^{'} (x))$
$\Rightarrow f (4) - f (2) = 2 \cdot (- 3)$
$\Rightarrow f (4) = - 6 + 10 = 4$

Q. Suppose f is a continuous function and f′(x) exists everywhere. If f(2)=10 and f′(x)≥−3 for all x, then the smallest possible value for f(4) is

1

2

3

4

For minimum value of f(4) 4−2f(4)−f(2)​=min(f′(x)) ⇒f(4)−f(2)=2⋅(−3) ⇒f(4)=−6+10=4

Q. Suppose $f$ is a continuous function and $f^{'} (x)$ exists everywhere. If $f (2) = 10$ and $f^{'} (x) \geq - 3$ for all $x$ , then the smallest possible value for $f (4)$ is

For minimum value of $f (4)$
$\frac{f ( 4 ) - f ( 2 )}{4 - 2} = min (f^{'} (x))$
$\Rightarrow f (4) - f (2) = 2 \cdot (- 3)$
$\Rightarrow f (4) = - 6 + 10 = 4$