Question

this problem is similar to the example in your textbook about guessing a formula for the derivative of f(x) = x² and to a similar problem in the exercises. study that example carefully before doing this problem. let f(x) = ⅓x³ + 10. estimate the following to within two decimal places by using small enough intervals. a. f(1) ≈ b. f(-5) ≈ c. f(8) ≈ d. determine a formula for f(x).

Explanation:

Step1: Recall the power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$, and the derivative of a constant $C$ is $0$. Given $f(x)=\frac{1}{3}x^{3}+10$.

Step2: Differentiate term - by - term

For the first term $\frac{1}{3}x^{3}$, using the power - rule with $a=\frac{1}{3}$ and $n = 3$, we have $\frac{d}{dx}(\frac{1}{3}x^{3})=\frac{1}{3}\times3x^{3 - 1}=x^{2}$. The derivative of the constant term $\frac{d}{dx}(10)=0$. So, $f^\prime(x)=x^{2}$.

Step3: Evaluate $f^\prime(x)$ at different points

a. When $x = 4$

Substitute $x = 4$ into $f^\prime(x)$: $f^\prime(4)=4^{2}=16.00$.

b. When $x=- 5$

Substitute $x=-5$ into $f^\prime(x)$: $f^\prime(-5)=(-5)^{2}=25.00$.

c. When $x = 8$

Substitute $x = 8$ into $f^\prime(x)$: $f^\prime(8)=8^{2}=64.00$.

d. Formula for $f^\prime(x)$

We already found that $f^\prime(x)=x^{2}$.

Answer:

a. $16.00$
b. $25.00$
c. $64.00$
d. $f^\prime(x)=x^{2}$