+0  
 
+1
262
4
avatar+322 

Evaluate and Simplify.

 

f(x) = 2 + 4x − x^2,        f[(5 + h) − f(5)] / h

 

I got an answer but I am not sure if I am right or not.  My answer was, "4 - h"

 Jan 16, 2019
edited by Ruublrr  Jan 17, 2019
 #1
avatar+533 
+1

i have seen this before, i will show you.

 

also to everyone else who is confused about the operations, it is \(\frac{f(5 + h)-f(5)}{h}\), not something else.

 

now, we have to calculate the numerator. f(h+5) is 2+4h+20-h^2-10h-25, and f(5) is -3. now you have (-h^2-6h)/h for the whole fraction, so the answer is -h-6. 

 

 

someone check my work pleez

 

and hope this helped!

 Jan 16, 2019
 #4
avatar+322 
+1

I am doing the problem again and I found my mistake. Thank you.

Ruublrr  Jan 17, 2019
 #2
avatar+533 
+1

ok here is my check: f(6)-f(5) should equal 3 according to you, 

 

according to me it should be -7.

 

plugging in f(6), it is 2+24-36 or -10, and f(5) is 2+20-25 or -3, so it is -7 and i am right.

 

hope this helped, and the answer is actually -h-6.

 Jan 16, 2019
 #3
avatar+106943 
+3

I'll check asad.

 

Evaluate and Simplify.

 

f(x) = 2 + 4x − x^2,        f(5 + h) − f(5) / h

 

I got an answer but I am not sure if I am right or not.  My answer was, "4 - h"

 

\(\frac{f(5+h)-f(5)}{h}\\ =\frac{[2+4(5+h)-(5+h)^2]-[2+20-25]}{h}\\ =\frac{[2+20+4h-(25+h^2+10h)]-[-3]}{h}\\ =\frac{2+20+4h-25-h^2-10h+3}{h}\\ =\frac{4h-h^2-10h}{h}\\ =\frac{-h^2-6h}{h}\\ =\frac{h(-h-6)}{h}\qquad \text{where } h>0\\ =-h-6\)

 

 

So Asad, you were correct.   Good work!   

 Jan 16, 2019

6 Online Users