# How do you find the instantaneous rate of change of y=4x^3+2x-3 at x=2?

Mar 25, 2015

The answer is: $50$.

The instantanous rate of change of the function $f \left(x\right)$ in $x = a$ is the slope of the tangent line to the graph of the function in that point:

$m = y ' \left(a\right)$.

So:

$y ' = 12 {x}^{2} + 2$ and:

$y ' \left(2\right) = 12 \cdot 4 + 2 = 50$.

Mar 25, 2015

Do you have rules for differentiation or are you using the definition? (a definition)

Important: You'll get the same answer either way, but I don't know where you are in your study of calculus.
If you are using a definition, it's probably one of the two below:

.

$f \left(x\right) = 4 {x}^{3} + 2 x - 3$

Solution 1:

Using definition: ${\lim}_{x \rightarrow 2} \frac{f \left(x\right) - f \left(2\right)}{x - 2}$

${\lim}_{x \rightarrow 2} \frac{f \left(x\right) - f \left(2\right)}{x - 2} = {\lim}_{x \rightarrow 2} \frac{\left[4 {x}^{3} + 2 x - 3\right] - \left[4 {\left(2\right)}^{3} + 2 \left(2\right) - 3\right]}{x - 2}$

$= {\lim}_{x \rightarrow 2} \frac{4 {x}^{3} + 2 x - 3 - 33}{x - 2} = {\lim}_{x \rightarrow 2} \frac{4 {x}^{3} + 2 x - 36}{x - 2}$

Trying to evaluate this limit by substitution gives indeterminate form: $\frac{0}{0}$. But don't give up hope!

Because $2$ is a zero of the polynomial numerator, we can be sure that $\left(x - 2\right)$ is a factor.

$4 {x}^{3} + 2 x - 36 = \left(x - 2\right) \left(4 {x}^{2} + 8 x + 18\right)$ (by division or by trial and error or, perhaps, by grouping)

Resuming:
${\lim}_{x \rightarrow 2} \frac{f \left(x\right) - f \left(2\right)}{x - 2} = {\lim}_{x \rightarrow 2} \frac{4 {x}^{3} + 2 x - 36}{x - 2}$

$= {\lim}_{x \rightarrow 2} \frac{\left(x - 2\right) \left(4 {x}^{2} + 8 x + 18\right)}{x - 2} = {\lim}_{x \rightarrow 2} \left(4 {x}^{2} + 8 x + 18\right)$

$= 4 {\left(2\right)}^{2} + 8 \left(2\right) + 18 = 16 + 16 + 18 = 50$

Solution 2:

Use the definition: ${\lim}_{h \rightarrow 0} \frac{f \left(2 + h\right) - f \left(2\right)}{h}$

${\lim}_{h \rightarrow 0} \frac{f \left(2 + h\right) - f \left(2\right)}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\left[4 {\left(2 + h\right)}^{3} + 2 \left(2 + h\right) - 3\right] - \left[4 {\left(2\right)}^{3} + 2 \left(2\right) - 3\right]}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\left[4 \left(8 + 12 h + 6 {h}^{2} + {h}^{3}\right) + 2 \left(2 + h\right) - 3\right] - \left[33\right]}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\left[32 + 48 h + 24 {h}^{2} + 4 {h}^{3} + 4 + 2 h - 3\right] - \left[33\right]}{h}$

$= {\lim}_{h \rightarrow 0} \frac{50 h + 24 {h}^{2} + 4 {h}^{3}}{h}$

$= {\lim}_{h \rightarrow 0} \left(50 + 24 h + 4 {h}^{2}\right) = 50$