Question

question 3
if f(x)=2x^3 - 13x^2 + 2x + 12 find f(x)
6x^2 - 26x + 2
6x^2 - 26x + 14
6x^3 - 26x^2 + 2x + 12
6x - 26
question 4
if f(x)=-\frac{1}{2}x^2+\frac{1}{x^3}, find f(1)

Explanation:

Step1: Recall power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For a polynomial function $f(x)=a_nx^n+a_{n - 1}x^{n - 1}+\cdots+a_1x + a_0$, $f^\prime(x)=na_nx^{n - 1}+(n - 1)a_{n - 1}x^{n - 2}+\cdots+a_1$.

Step2: Differentiate $f(x)=2x^3-13x^2 + 2x+12$ term - by - term

For the term $2x^3$, using the power - rule, its derivative is $3\times2x^{3 - 1}=6x^2$. For the term $-13x^2$, its derivative is $2\times(-13)x^{2 - 1}=-26x$. For the term $2x$, its derivative is $1\times2x^{1 - 1}=2$. The derivative of the constant term $12$ is $0$. So, $f^\prime(x)=6x^2-26x + 2$.

Step3: For $f(x)=-\frac{1}{2}x^2+\frac{1}{x^3}=-\frac{1}{2}x^2+x^{-3}$

Differentiate term - by - term. The derivative of $-\frac{1}{2}x^2$ is $2\times(-\frac{1}{2})x^{2 - 1}=-x$. The derivative of $x^{-3}$ is $-3x^{-3 - 1}=-3x^{-4}=-\frac{3}{x^4}$. So, $f^\prime(x)=-x-\frac{3}{x^4}$.

Step4: Evaluate $f^\prime(x)$ at $x = 1$

Substitute $x = 1$ into $f^\prime(x)=-x-\frac{3}{x^4}$. We get $f^\prime(1)=-1-\frac{3}{1^4}=-1 - 3=-4$.

Answer:

Question 3: A. $6x^2-26x + 2$
Question 4: $-4$