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Let f(x) = (3x² + x − 5) / (x² + 4). Find f(3), f(2), f(1), f(a+1), f(sqrt(a)), f(a²), f(a) + f(1).

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Answers

4 parts
a

a2+b2+2ab+2a+2b−4a^2 + b^2 + 2ab + 2a + 2b - 4

b

a2+b2−2ab+2a−2b−4a^2 + b^2 - 2ab + 2a - 2b - 4

c

b+4

d

2a + b + 2

Graph of f(x) = (3x² + x − 5)/(x² + 4)

Use the left and right arrow keys to trace the curve, up and down to jump between key points, and plus and minus to zoom.

Key points of f(x) = (3x² + x − 5)/(x² + 4)
Pointxy
x-intercept−1.470
Minimum−0.12−1.26
y-intercept0−1.25
x-intercept1.140

Step-by-step solution

3 steps

Step 1: Understand the given function.
We are given the function h(x)=x2+2x−4h(x) = x^2 + 2x - 4.

Step 2: Solve part a).

Part a· 1 step

Find h(b+a)h(b+a).

Substitute x=(b+a)x = (b+a) into the function:

h(b+a)=(b+a)2+2(b+a)−4h(b+a) = (b+a)^2 + 2(b+a) - 4

Expand the terms:

h(b+a)=(b2+2ab+a2)+(2b+2a)−4h(b+a) = (b^2 + 2ab + a^2) + (2b + 2a) - 4 h(b+a)=a2+b2+2ab+2a+2b−4h(b+a) = a^2 + b^2 + 2ab + 2a + 2b - 4

The simplified expression is a2+b2+2ab+2a+2b−4\boxed{a^2 + b^2 + 2ab + 2a + 2b - 4}.

  1. Solve part b).

Part b

a2+b2−2ab+2a−2b−4a^2 + b^2 - 2ab + 2a - 2b - 4

Find h(a−b)h(a-b).

Substitute x=(a−b)x = (a-b) into the function:

h(a−b)=(a−b)2+2(a−b)−4h(a-b) = (a-b)^2 + 2(a-b) - 4

Expand the terms:

h(a−b)=(a2−2ab+b2)+(2a−2b)−4h(a-b) = (a^2 - 2ab + b^2) + (2a - 2b) - 4 h(a−b)=a2+b2−2ab+2a−2b−4h(a-b) = a^2 + b^2 - 2ab + 2a - 2b - 4

The simplified expression is a2+b2−2ab+2a−2b−4\boxed{a^2 + b^2 - 2ab + 2a - 2b - 4}.

  1. Solve part c).

Part cb+4

Find h(1+b)−h(1)b\frac{h(1+b)-h(1)}{b}.

First, find h(1+b)h(1+b):

h(1+b)=(1+b)2+2(1+b)−4h(1+b) = (1+b)^2 + 2(1+b) - 4 h(1+b)=(1+2b+b2)+(2+2b)−4h(1+b) = (1 + 2b + b^2) + (2 + 2b) - 4 h(1+b)=b2+4b+3−4h(1+b) = b^2 + 4b + 3 - 4 h(1+b)=b2+4b−1h(1+b) = b^2 + 4b - 1

Next, find h(1)h(1):

h(1)=(1)2+2(1)−4h(1) = (1)^2 + 2(1) - 4 h(1)=1+2−4h(1) = 1 + 2 - 4 h(1)=−1h(1) = -1

Now, substitute these into the expression:

h(1+b)−h(1)b=(b2+4b−1)−(−1)b\frac{h(1+b)-h(1)}{b} = \frac{(b^2 + 4b - 1) - (-1)}{b} h(1+b)−h(1)b=b2+4b−1+1b\frac{h(1+b)-h(1)}{b} = \frac{b^2 + 4b - 1 + 1}{b} h(1+b)−h(1)b=b2+4bb\frac{h(1+b)-h(1)}{b} = \frac{b^2 + 4b}{b}

Factor out bb from the numerator:

h(1+b)−h(1)b=b(b+4)b\frac{h(1+b)-h(1)}{b} = \frac{b(b+4)}{b}

Cancel bb (assuming b≠0b \neq 0):

h(1+b)−h(1)b=b+4\frac{h(1+b)-h(1)}{b} = b+4

The simplified expression is b+4\boxed{b+4}.

  1. Solve part d).

Part d2a + b + 2

Find h(a+b)−h(a)b\frac{h(a+b)-h(a)}{b}.

First, find h(a+b)h(a+b):

h(a+b)=(a+b)2+2(a+b)−4h(a+b) = (a+b)^2 + 2(a+b) - 4 h(a+b)=a2+2ab+b2+2a+2b−4h(a+b) = a^2 + 2ab + b^2 + 2a + 2b - 4

Next, find h(a)h(a):

h(a)=a2+2a−4h(a) = a^2 + 2a - 4

Now, substitute these into the expression:

h(a+b)−h(a)b=(a2+2ab+b2+2a+2b−4)−(a2+2a−4)b\frac{h(a+b)-h(a)}{b} = \frac{(a^2 + 2ab + b^2 + 2a + 2b - 4) - (a^2 + 2a - 4)}{b}

Distribute the negative sign in the numerator:

h(a+b)−h(a)b=a2+2ab+b2+2a+2b−4−a2−2a+4b\frac{h(a+b)-h(a)}{b} = \frac{a^2 + 2ab + b^2 + 2a + 2b - 4 - a^2 - 2a + 4}{b}

Combine like terms in the numerator:

h(a+b)−h(a)b=(a2−a2)+(2a−2a)+(−4+4)+2ab+b2+2bb\frac{h(a+b)-h(a)}{b} = \frac{(a^2 - a^2) + (2a - 2a) + (-4 + 4) + 2ab + b^2 + 2b}{b} h(a+b)−h(a)b=2ab+b2+2bb\frac{h(a+b)-h(a)}{b} = \frac{2ab + b^2 + 2b}{b}

Factor out bb from the numerator:

h(a+b)−h(a)b=b(2a+b+2)b\frac{h(a+b)-h(a)}{b} = \frac{b(2a + b + 2)}{b}

Cancel bb (assuming b≠0b \neq 0):

h(a+b)−h(a)b=2a+b+2\frac{h(a+b)-h(a)}{b} = 2a + b + 2

The simplified expression is 2a+b+2\boxed{2a + b + 2}.

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