Question

attempt 1: 2 out of 6 parts have been answered correctly. (this problem is similar to the example in your textbook about guessing a formula for the derivative of $f(x)=x^{2}$ and to a similar problem in the exercises. study that example carefully before doing this problem.) let $f(x)=x^{3}+9$. estimate the following to within two decimal places by using small enough intervals. a. $f(4)approx$ b. $f(-3)approx$ c. $f(6)approx$ d. $f(7)approx$ e. $f(0)approx$

Explanation:

Step1: Recall derivative formula

The derivative of $f(x)=x^{3}+9$ using the power - rule $\frac{d}{dx}(x^{n}) = nx^{n - 1}$ and $\frac{d}{dx}(c)=0$ (where $c$ is a constant) is $f^{\prime}(x)=3x^{2}$.

Step2: Calculate $f^{\prime}(4)$

Substitute $x = 4$ into $f^{\prime}(x)$: $f^{\prime}(4)=3\times4^{2}=3\times16 = 48.00$.

Step3: Calculate $f^{\prime}(-3)$

Substitute $x=-3$ into $f^{\prime}(x)$: $f^{\prime}(-3)=3\times(-3)^{2}=3\times9 = 27.00$.

Step4: Calculate $f^{\prime}(6)$

Substitute $x = 6$ into $f^{\prime}(x)$: $f^{\prime}(6)=3\times6^{2}=3\times36 = 108.00$.

Step5: Calculate $f^{\prime}(7)$

Substitute $x = 7$ into $f^{\prime}(x)$: $f^{\prime}(7)=3\times7^{2}=3\times49 = 147.00$.

Step6: Calculate $f^{\prime}(0)$

Substitute $x = 0$ into $f^{\prime}(x)$: $f^{\prime}(0)=3\times0^{2}=0.00$.

Answer:

a. $f^{\prime}(4)=48.00$
b. $f^{\prime}(-3)=27.00$
c. $f^{\prime}(6)=108.00$
d. $f^{\prime}(7)=147.00$
e. $f^{\prime}(0)=0.00$