Question

attempt 2: 9 attempts remaining. let ( f(x) = 3x^2 + 5x + 5 ) and let ( g(h) = \frac{f(3 + h) - f(3)}{h} ). determine each of the following: (a) ( g(1) = ) (b) ( g(0.1) = ) (c) ( g(0.01) = ) you will notice that the values that you entered are getting closer and closer to a number ( l ). this number is called the limit of ( g(h) ) as ( h ) approaches 0 and is also called the derivative of ( f(x) ) at the point when ( x = 3 ). we will see more of this when we get to the calculus textbook. enter the value of ( l ): video example: solving a similar problem

Explanation:

Part (a): Find \( g(1) \)

Step 1: Compute \( f(3 + h) \) when \( h = 1 \)

First, find \( f(3 + 1)=f(4) \). Given \( f(x)=3x^{2}+5x + 5 \), substitute \( x = 4 \):
\( f(4)=3(4)^{2}+5(4)+5=3\times16 + 20+5=48 + 20 + 5=73 \)

Step 2: Compute \( f(3) \)

Substitute \( x = 3 \) into \( f(x) \):
\( f(3)=3(3)^{2}+5(3)+5=3\times9+15 + 5=27+15 + 5=47 \)

Step 3: Compute \( g(1) \)

Using the formula \( g(h)=\frac{f(3 + h)-f(3)}{h} \), with \( h = 1 \):
\( g(1)=\frac{f(4)-f(3)}{1}=\frac{73 - 47}{1}=26 \)

Part (b): Find \( g(0.1) \)

Step 1: Compute \( f(3 + 0.1)=f(3.1) \)

Substitute \( x = 3.1 \) into \( f(x) \):
\( f(3.1)=3(3.1)^{2}+5(3.1)+5=3\times9.61+15.5 + 5=28.83+15.5 + 5=49.33 \)

Step 2: Use \( f(3)=47 \) (from part a)

Step 3: Compute \( g(0.1) \)

\( g(0.1)=\frac{f(3.1)-f(3)}{0.1}=\frac{49.33 - 47}{0.1}=\frac{2.33}{0.1}=23.3 \)

Part (c): Find \( g(0.01) \)

Step 1: Compute \( f(3 + 0.01)=f(3.01) \)

Substitute \( x = 3.01 \) into \( f(x) \):
\( f(3.01)=3(3.01)^{2}+5(3.01)+5=3\times9.0601+15.05 + 5=27.1803+15.05 + 5=47.2303 \)

Step 2: Use \( f(3)=47 \) (from part a)

Step 3: Compute \( g(0.01) \)

\( g(0.01)=\frac{f(3.01)-f(3)}{0.01}=\frac{47.2303 - 47}{0.01}=\frac{0.2303}{0.01}=23.03 \)

Part (d): Find the limit \( L \) as \( h

ightarrow0 \)
The derivative of \( f(x) \) at \( x = a \) is given by \( f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h} \). For \( f(x)=3x^{2}+5x + 5 \), the derivative \( f^{\prime}(x)=6x + 5 \) (using the power rule: \( \frac{d}{dx}(x^{n})=nx^{n - 1} \), so \( \frac{d}{dx}(3x^{2}) = 6x \), \( \frac{d}{dx}(5x)=5 \), \( \frac{d}{dx}(5)=0 \)).
Substitute \( x = 3 \) into \( f^{\prime}(x) \):
\( f^{\prime}(3)=6(3)+5=18 + 5=23 \)
So the limit \( L = 23 \)

Answer:

s:
(a) \( \boldsymbol{26} \)

(b) \( \boldsymbol{23.3} \)

(c) \( \boldsymbol{23.03} \)

(d) \( \boldsymbol{23} \)