use the limit definition of the derivative to find the instantaneous rate of change of $f(x)=4x^{2}+6x + 6$…

use the limit definition of the derivative to find the instantaneous rate of change of $f(x)=4x^{2}+6x + 6$ at $x = 4$. question help: video message instructor submit question jump to answer

use the limit definition of the derivative to find the instantaneous rate of change of $f(x)=4x^{2}+6x + 6$ at $x = 4$. question help: video message instructor submit question jump to answer

Answer

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. First, find $f(x + h)$ for $f(x)=4x^{2}+6x + 6$. $f(x + h)=4(x + h)^{2}+6(x + h)+6=4(x^{2}+2xh+h^{2})+6x+6h + 6=4x^{2}+8xh+4h^{2}+6x+6h + 6$.

Step2: Calculate $f(x + h)-f(x)$

$f(x + h)-f(x)=(4x^{2}+8xh+4h^{2}+6x+6h + 6)-(4x^{2}+6x + 6)=8xh+4h^{2}+6h$.

Step3: Calculate $\frac{f(x + h)-f(x)}{h}$

$\frac{f(x + h)-f(x)}{h}=\frac{8xh+4h^{2}+6h}{h}=8x + 4h+6$.

Step4: Find the derivative $f^\prime(x)$

$f^\prime(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h\rightarrow0}(8x + 4h+6)$. As $h\rightarrow0$, $f^\prime(x)=8x+6$.

Step5: Evaluate the derivative at $x = 4$

Substitute $x = 4$ into $f^\prime(x)$. $f^\prime(4)=8\times4+6=32 + 6=38$.

Answer:

38