Solved on Jan 29, 2024

Use $f(x)=4x^2$ to complete the table. Enter the answer as a whole number, fraction, or decimal rounded to 2 decimals. For fractions, use $\frac{1}{2}$ or $-\frac{1}{2}$ format.

STEP 1

Assumptions
1. The function given is $f(x) = 4x^2$ .
2. We need to calculate the value of the expression $\frac{f(x+h)-f(x)}{h}$ for various values of $h$ when $x = 5$ .
3. The results should be expressed as whole numbers, fractions, or decimals rounded to two decimal places.

STEP 2

First, let's write down the general form of the expression we need to calculate.
$\frac{f(x+h)-f(x)}{h} = \frac{4(x+h)^2 - 4x^2}{h}$

STEP 3

Now, let's expand the expression $(x+h)^2$ .
$(x+h)^2 = x^2 + 2xh + h^2$

STEP 4

Substitute the expanded form back into the expression.
$\frac{f(x+h)-f(x)}{h} = \frac{4(x^2 + 2xh + h^2) - 4x^2}{h}$

STEP 5

Distribute the 4 across the terms inside the parentheses.
$\frac{f(x+h)-f(x)}{h} = \frac{4x^2 + 8xh + 4h^2 - 4x^2}{h}$

STEP 6

Notice that $4x^2$ and $-4x^2$ cancel each other out.
$\frac{f(x+h)-f(x)}{h} = \frac{8xh + 4h^2}{h}$

STEP 7

Factor out $h$ from the numerator.
$\frac{f(x+h)-f(x)}{h} = \frac{h(8x + 4h)}{h}$

STEP 8

Cancel out $h$ from the numerator and denominator.
$\frac{f(x+h)-f(x)}{h} = 8x + 4h$

STEP 9

Now we will evaluate this expression for $x = 5$ and various values of $h$ .

STEP 10

First, let's calculate it for $h = 2$ .
$\frac{f(5+2)-f(5)}{2} = 8 \cdot 5 + 4 \cdot 2$

STEP 11

Perform the multiplication.
$\frac{f(5+2)-f(5)}{2} = 40 + 8$

STEP 12

Add the results.
$\frac{f(5+2)-f(5)}{2} = 48$
This matches the value given in the table for $h = 2$ .

STEP 13

Next, calculate it for $h = 1$ .
$\frac{f(5+1)-f(5)}{1} = 8 \cdot 5 + 4 \cdot 1$

STEP 14

Perform the multiplication.
$\frac{f(5+1)-f(5)}{1} = 40 + 4$

STEP 15

Add the results.
$\frac{f(5+1)-f(5)}{1} = 44$
This matches the value given in the table for $h = 1$ .

STEP 16

Now, calculate it for $h = 0.1$ .
$\frac{f(5+0.1)-f(5)}{0.1} = 8 \cdot 5 + 4 \cdot 0.1$

STEP 17

Perform the multiplication.
$\frac{f(5+0.1)-f(5)}{0.1} = 40 + 0.4$

STEP 18

Add the results.
$\frac{f(5+0.1)-f(5)}{0.1} = 40.4$
This matches the value given in the table for $h = 0.1$ .

STEP 19

Finally, calculate it for $h = 0.01$ .
$\frac{f(5+0.01)-f(5)}{0.01} = 8 \cdot 5 + 4 \cdot 0.01$

STEP 20

Perform the multiplication.
$\frac{f(5+0.01)-f(5)}{0.01} = 40 + 0.04$

STEP 21

Add the results.
$\frac{f(5+0.01)-f(5)}{0.01} = 40.04$
This matches the value given in the table for $h = 0.01$ .

STEP 22

Now we have verified all the given values in the table and found that they are correct.
The completed table with the expression $\frac{f(x+h)-f(x)}{h}$ evaluated for $x = 5$ and various $h$ is:
\begin{tabular}{|c|c|c|} \hline $\mathbf{x}$ & $\boldsymbol{h}$ & $\frac{f(x+h)-f(x)}{h}$ \\ \hline 5 & 2 & 48 \\ \hline 5 & 1 & 44 \\ \hline 5 & 0.1 & 40.4 \\ \hline 5 & 0.01 & 40.04 \\ \hline \end{tabular}

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Use f(x)=4x2f(x)=4x^2f(x)=4x2 to complete the table. Enter the answer as a whole number, fraction, or decimal rounded to 2 decimals. For fractions, use 12\frac{1}{2}21​ or −12-\frac{1}{2}−21​ format.

STEP 1

STEP 2

First, let's write down the general form of the expression we need to calculate.f(x+h)−f(x)h=4(x+h)2−4x2h \frac{f(x+h)-f(x)}{h} = \frac{4(x+h)^2 - 4x^2}{h} hf(x+h)−f(x)​=h4(x+h)2−4x2​

STEP 3

Now, let's expand the expression (x+h)2 (x+h)^2 (x+h)2.(x+h)2=x2+2xh+h2 (x+h)^2 = x^2 + 2xh + h^2 (x+h)2=x2+2xh+h2

STEP 4

Substitute the expanded form back into the expression.f(x+h)−f(x)h=4(x2+2xh+h2)−4x2h \frac{f(x+h)-f(x)}{h} = \frac{4(x^2 + 2xh + h^2) - 4x^2}{h} hf(x+h)−f(x)​=h4(x2+2xh+h2)−4x2​

STEP 5

Distribute the 4 across the terms inside the parentheses.f(x+h)−f(x)h=4x2+8xh+4h2−4x2h \frac{f(x+h)-f(x)}{h} = \frac{4x^2 + 8xh + 4h^2 - 4x^2}{h} hf(x+h)−f(x)​=h4x2+8xh+4h2−4x2​

STEP 6

Notice that 4x2 4x^2 4x2 and −4x2 -4x^2 −4x2 cancel each other out.f(x+h)−f(x)h=8xh+4h2h \frac{f(x+h)-f(x)}{h} = \frac{8xh + 4h^2}{h} hf(x+h)−f(x)​=h8xh+4h2​

STEP 7

Factor out h h h from the numerator.f(x+h)−f(x)h=h(8x+4h)h \frac{f(x+h)-f(x)}{h} = \frac{h(8x + 4h)}{h} hf(x+h)−f(x)​=hh(8x+4h)​

STEP 8

Cancel out h h h from the numerator and denominator.f(x+h)−f(x)h=8x+4h \frac{f(x+h)-f(x)}{h} = 8x + 4h hf(x+h)−f(x)​=8x+4h

STEP 9

Now we will evaluate this expression for x=5 x = 5 x=5 and various values of h h h.

STEP 10

First, let's calculate it for h=2 h = 2 h=2.f(5+2)−f(5)2=8⋅5+4⋅2 \frac{f(5+2)-f(5)}{2} = 8 \cdot 5 + 4 \cdot 2 2f(5+2)−f(5)​=8⋅5+4⋅2

STEP 11

Perform the multiplication.f(5+2)−f(5)2=40+8 \frac{f(5+2)-f(5)}{2} = 40 + 8 2f(5+2)−f(5)​=40+8

STEP 12

Add the results.f(5+2)−f(5)2=48 \frac{f(5+2)-f(5)}{2} = 48 2f(5+2)−f(5)​=48This matches the value given in the table for h=2 h = 2 h=2.

STEP 13

Next, calculate it for h=1 h = 1 h=1.f(5+1)−f(5)1=8⋅5+4⋅1 \frac{f(5+1)-f(5)}{1} = 8 \cdot 5 + 4 \cdot 1 1f(5+1)−f(5)​=8⋅5+4⋅1

STEP 14

Perform the multiplication.f(5+1)−f(5)1=40+4 \frac{f(5+1)-f(5)}{1} = 40 + 4 1f(5+1)−f(5)​=40+4

STEP 15

Add the results.f(5+1)−f(5)1=44 \frac{f(5+1)-f(5)}{1} = 44 1f(5+1)−f(5)​=44This matches the value given in the table for h=1 h = 1 h=1.

STEP 16

Now, calculate it for h=0.1 h = 0.1 h=0.1.f(5+0.1)−f(5)0.1=8⋅5+4⋅0.1 \frac{f(5+0.1)-f(5)}{0.1} = 8 \cdot 5 + 4 \cdot 0.1 0.1f(5+0.1)−f(5)​=8⋅5+4⋅0.1

STEP 17

Perform the multiplication.f(5+0.1)−f(5)0.1=40+0.4 \frac{f(5+0.1)-f(5)}{0.1} = 40 + 0.4 0.1f(5+0.1)−f(5)​=40+0.4

STEP 18

Add the results.f(5+0.1)−f(5)0.1=40.4 \frac{f(5+0.1)-f(5)}{0.1} = 40.4 0.1f(5+0.1)−f(5)​=40.4This matches the value given in the table for h=0.1 h = 0.1 h=0.1.

STEP 19

Finally, calculate it for h=0.01 h = 0.01 h=0.01.f(5+0.01)−f(5)0.01=8⋅5+4⋅0.01 \frac{f(5+0.01)-f(5)}{0.01} = 8 \cdot 5 + 4 \cdot 0.01 0.01f(5+0.01)−f(5)​=8⋅5+4⋅0.01

STEP 20

Perform the multiplication.f(5+0.01)−f(5)0.01=40+0.04 \frac{f(5+0.01)-f(5)}{0.01} = 40 + 0.04 0.01f(5+0.01)−f(5)​=40+0.04

STEP 21

Add the results.f(5+0.01)−f(5)0.01=40.04 \frac{f(5+0.01)-f(5)}{0.01} = 40.04 0.01f(5+0.01)−f(5)​=40.04This matches the value given in the table for h=0.01 h = 0.01 h=0.01.

STEP 22

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Use $f(x)=4x^2$ to complete the table. Enter the answer as a whole number, fraction, or decimal rounded to 2 decimals. For fractions, use $\frac{1}{2}$ or $-\frac{1}{2}$ format.

First, let's write down the general form of the expression we need to calculate.
$\frac{f(x+h)-f(x)}{h} = \frac{4(x+h)^2 - 4x^2}{h}$

Now, let's expand the expression $(x+h)^2$ .
$(x+h)^2 = x^2 + 2xh + h^2$

Substitute the expanded form back into the expression.
$\frac{f(x+h)-f(x)}{h} = \frac{4(x^2 + 2xh + h^2) - 4x^2}{h}$

Distribute the 4 across the terms inside the parentheses.
$\frac{f(x+h)-f(x)}{h} = \frac{4x^2 + 8xh + 4h^2 - 4x^2}{h}$

Notice that $4x^2$ and $-4x^2$ cancel each other out.
$\frac{f(x+h)-f(x)}{h} = \frac{8xh + 4h^2}{h}$

Factor out $h$ from the numerator.
$\frac{f(x+h)-f(x)}{h} = \frac{h(8x + 4h)}{h}$

Cancel out $h$ from the numerator and denominator.
$\frac{f(x+h)-f(x)}{h} = 8x + 4h$

Now we will evaluate this expression for $x = 5$ and various values of $h$ .

First, let's calculate it for $h = 2$ .
$\frac{f(5+2)-f(5)}{2} = 8 \cdot 5 + 4 \cdot 2$

Perform the multiplication.
$\frac{f(5+2)-f(5)}{2} = 40 + 8$

Add the results.
$\frac{f(5+2)-f(5)}{2} = 48$
This matches the value given in the table for $h = 2$ .

Next, calculate it for $h = 1$ .
$\frac{f(5+1)-f(5)}{1} = 8 \cdot 5 + 4 \cdot 1$

Perform the multiplication.
$\frac{f(5+1)-f(5)}{1} = 40 + 4$

Add the results.
$\frac{f(5+1)-f(5)}{1} = 44$
This matches the value given in the table for $h = 1$ .

Now, calculate it for $h = 0.1$ .
$\frac{f(5+0.1)-f(5)}{0.1} = 8 \cdot 5 + 4 \cdot 0.1$

Perform the multiplication.
$\frac{f(5+0.1)-f(5)}{0.1} = 40 + 0.4$

Add the results.
$\frac{f(5+0.1)-f(5)}{0.1} = 40.4$
This matches the value given in the table for $h = 0.1$ .

Finally, calculate it for $h = 0.01$ .
$\frac{f(5+0.01)-f(5)}{0.01} = 8 \cdot 5 + 4 \cdot 0.01$

Perform the multiplication.
$\frac{f(5+0.01)-f(5)}{0.01} = 40 + 0.04$

Add the results.
$\frac{f(5+0.01)-f(5)}{0.01} = 40.04$
This matches the value given in the table for $h = 0.01$ .