If f(x) = x2 - 2x + 4 on [1, 5], then the value of a constant c such that (f(5) - f(1)/5 - 1) = f'(c), is:

Question

If f(x) = x2 - 2x + 4 on [1, 5], then the value of a constant c such that (f(5) - f(1)/5 - 1) = f'(c), is: (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Given f (x) = x2 - 2x + 4 ∴ f'(x) = 2x - 2 So, f'(c) = 2c - 2 So, 2c - 2 = (f(5) - f(1)/5 - 1) ⇒ f(5) = (5)2 - 2 (5) + 4 = 19 and f(1) = (1)2 - 2(1) + 4 = 3 ⇒ 2c - 2 = (19 - 3/4) = 4 ⇒ 2c - 6 Thus c = 3.

Q. If $f (x) = x^{2} - 2 x + 4$ on [1, 5], then the value of a constant c such that $\frac{f ( 5 ) - f ( 1 )}{5 - 1} = f^{'} (c)$ , is:

0

1

2

3

Given $f (x) = x^{2} - 2 x + 4$
$∴ f^{'} (x) = 2 x - 2$
So, $f^{'} (c) = 2 c - 2$
So, $2 c - 2 = \frac{f ( 5 ) - f ( 1 )}{5 - 1}$
$\Rightarrow f (5) = (5)^{2} - 2 (5) + 4 = 19$ and $f (1) = (1)^{2} - 2 (1) + 4 = 3$
$\Rightarrow 2 c - 2 = \frac{19 - 3}{4} = 4 \Rightarrow 2 c - 6$
Thus $c = 3.$

Q. If f(x)=x2−2x+4 on [1, 5], then the value of a constant c such that 5−1f(5)−f(1)​=f′(c), is:

0

1

2

3

Given f(x)=x2−2x+4 ∴f′(x)=2x−2 So, f′(c)=2c−2 So, 2c−2=5−1f(5)−f(1)​ ⇒f(5)=(5)2−2(5)+4=19 and f(1)=(1)2−2(1)+4=3 ⇒2c−2=419−3​=4⇒2c−6 Thus c=3.

Q. If $f (x) = x^{2} - 2 x + 4$ on [1, 5], then the value of a constant c such that $\frac{f ( 5 ) - f ( 1 )}{5 - 1} = f^{'} (c)$ , is:

Given $f (x) = x^{2} - 2 x + 4$
$∴ f^{'} (x) = 2 x - 2$
So, $f^{'} (c) = 2 c - 2$
So, $2 c - 2 = \frac{f ( 5 ) - f ( 1 )}{5 - 1}$
$\Rightarrow f (5) = (5)^{2} - 2 (5) + 4 = 19$ and $f (1) = (1)^{2} - 2 (1) + 4 = 3$
$\Rightarrow 2 c - 2 = \frac{19 - 3}{4} = 4 \Rightarrow 2 c - 6$
Thus $c = 3.$