Question

1. $c(x) = 0.5x^2 + 8x$ use the limit definition: $c(20) = \lim\limits_{h \to 0} \frac{c(20 + h) - c(20)}{h}$ show all steps.

Explanation:

Step1: Find \( C(20 + h) \)

Substitute \( x = 20 + h \) into \( C(x)=0.5x^{2}+8x \):
\( C(20 + h)=0.5(20 + h)^{2}+8(20 + h) \)
Expand \( (20 + h)^{2}=400 + 40h + h^{2} \):
\( C(20 + h)=0.5(400 + 40h + h^{2})+160 + 8h \)
\( = 200 + 20h + 0.5h^{2}+160 + 8h \)
\( = 0.5h^{2}+28h + 360 \)

Step2: Find \( C(20) \)

Substitute \( x = 20 \) into \( C(x)=0.5x^{2}+8x \):
\( C(20)=0.5(20)^{2}+8(20) \)
\( = 0.5\times400 + 160 \)
\( = 200 + 160 = 360 \)

Step3: Compute \( C(20 + h)-C(20) \)

Subtract \( C(20) \) from \( C(20 + h) \):
\( C(20 + h)-C(20)=(0.5h^{2}+28h + 360)-360 \)
\( = 0.5h^{2}+28h \)

Step4: Compute the limit

Now, find \( \lim_{h
ightarrow0}\frac{C(20 + h)-C(20)}{h} \):
\( \lim_{h
ightarrow0}\frac{0.5h^{2}+28h}{h} \)
Factor out \( h \) from the numerator:
\( \lim_{h
ightarrow0}\frac{h(0.5h + 28)}{h} \)
Cancel out \( h \) (for \( h
eq0 \)):
\( \lim_{h
ightarrow0}(0.5h + 28) \)
Substitute \( h = 0 \):
\( = 0.5\times0 + 28 = 28 \)

Answer:

\( C'(20)=28 \)