Question

For the function f defined by $f\left(x\right)=x^2-4x+1$, find:

(a) f(2)

(b) f(x) + f(2)

(c) f(-x)

(d) -f(x)

(e) f(x + 2)

(f)$\frac{f(x+h)-f\left(x\right)}{h},h\ne 0$

Short answer

(a)$f\left(2\right)=-3$

(b) $f\left(x\right)+f\left(2\right)=x^2-4x-2$

(c) $f(-x)=x^2+4x+1$

(d) $-f\left(x\right)=-x^2+4x-1$

(e) $f(x+2)=x^2-3$

(f)$\frac{f(x+h)-f\left(x\right)}{h}=2x+h-4,h\ne 0$

Step-by-step solution

Part (a) Step 1. Given Information

$f\left(x\right)=x^2-4x+1$

Part (a) Step 2. Calculation

Now, $f\left(2\right)=2^2-4.2+1$

$=-3$

Part (b) Step 1. Given Information

$f\left(x\right)=x^2-4x+1$

Part (b) Step 2. Calculation

$f\left(x\right)+f\left(2\right)=x^2-4x+1-3$

$=x^2-4x-2$

Part (c) Step 1. Given Information 

$f\left(x\right)=x^2-4x+1$

Part (c) Step 2. Calculation 

$f(-x)={(-x)}^2-4(-x)+1$

role="math" localid="1647455449353" $=x^2+4x+1$

Part (d) Step 1. Given Information 

$f\left(x\right)=x^2-4x+1$

Part (d) Step 2. Calculation

$-f\left(x\right)=-x^2+4x-1$

Part (e) Step 1. Given Information

$f\left(x\right)=x^2-4x+1$

Part (e) Step 2. Calculation

$f(x+2)={(x+2)}^2-4(x+2)+1$

$=x^2+4x+4-4x-8+1$

$=x^2-3$

Part (f) Step 1. Given Information 

$f\left(x\right)=x^2-4x+1$

Part (f) Step 2.Calculation

$\frac{f(x+h)-f\left(x\right)}{h}=\frac{\{{(x+h)}^2-4(x+h)+1\}-\{x^2-4x+1\}}{h},h\ne 0$

$=\frac{x^2+2xh+h^2-4x-4h+1-x^2+4x-1}{h}$

$=\frac{h(2x-4+h)}{h}$

$=2x+h-4$