Question

Consider the function f(x)=x^3-4x^2+2. Calculate the limit of the difference quotient at x0=3 for f(x).

Answer 1

The limit of the difference quotient of the above function $f(x)$ at $x=3$ is $3$ such that $f(x)=x^{3} - 4x^{2} + 2$.

Difference of quotient

The difference quotient of a function $f(x)$ is $\frac{f(x+h)-f(x)}{h}$.

How to evaluate the limit of the function?

The given equation is, $f(x)=x^{3} -4x^{2} +2$

So, $f(x+h)=(x+h)^{3} -4(x+h)^{2} +2= x^{3} +h^{3}+3x^{2} h+3xh^{2} -4x^{2} -4h^{2} -8xh+2$

Now, $f(x+h)-f(x)$

$=x^{3}+h^{3}+3x^{2}h+3xh^{2}-4x^{2}-4h^{2}-8xh+2-x^{3}+4x^{2}-2$

$=h^{3}+3x^{2}h+3xh^{2}-4h^{2}-8xh$

So, $\frac{f(x+h)-f(x)}{h} =\frac{h^{3}+3x^{2}h+3xh^{2} -4h^{2}-8xh }{h}$

$=h^{2}+3x^{2}+3xh-4h-8x$

Now, at $x=3$,

$h^{2}+3x^{2}+3xh-4h-8x=h^{2}+27+9h-4h-24=h^{2}+5h+3$

If   $h$→$0$, the value of $h^{2}+5h+3=3$

Thus, the limit of the difference quotient of the above function $f(x)$ at $x=3$ is $3$.

Learn more about the limit of the difference quotient here- https://brainly.com/question/17008881

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