Let f(x)=(x-2)17(x+5)24. Then

Question

Let f(x)=(x-2)17(x+5)24. Then (A) f does not have a critical point at x=2 (B) f has a minimum at x=2 (C) f has neither a maximum nor a minimum at x=2 (D) f has a maximum at x=2

Tardigrade Admin · Accepted Answer

f(x)=(x-2)17(x+5)24 ⇒ f prime(x)=17(x-2)16(x+5)24+24(x-2)17(x+5)23 =(x-2)16(x+5)23(17 x+85+24 x-48)=(x-2)

Q. Let $f (x) = (x - 2)^{17} (x + 5)^{24}$ . Then

$f$ does not have a critical point at $x = 2$

$f$ has a minimum at $x = 2$

$f$ has neither a maximum nor a minimum at $x = 2$

$f$ has a maximum at $x = 2$

$f (x) = (x - 2)^{17} (x + 5)^{24}$
$\Rightarrow f^{'} (x) = 17 (x - 2)^{16} (x + 5)^{24} + 24 (x - 2)^{17} (x + 5)^{23}$
$= (x - 2)^{16} (x + 5)^{23} (17 x + 85 + 24 x - 48) = (x - 2)$

Q. Let f(x)=(x−2)17(x+5)24. Then

f does not have a critical point at x=2

f has a minimum at x=2

f has neither a maximum nor a minimum at x=2

f has a maximum at x=2