# f(x)=a(x^{2}) −8x lim f(x)=-1 xrightarrow2

Question
Functions
$$f(x)=a(x^{2}) −8x$$
$$lim f(x)=-1$$
$$x\rightarrow2$$

2021-02-04
$$lim f(x)=-1$$ Given $$x\rightarrow2$$
$$lim ax^{2}-8x=-1$$ Substitute f(x)=ax^{2}-8x\) $$x\rightarrow2$$
$$a2^{2}−8(2)=-1$$ Evaluate the limit by substituting x=2.
$$4a-16=-1$$ Simplify.
$$4a=15$$ Add 16 on both sides.
$$a=\frac{15}{4}$$ Divide both sides by 4.

### Relevant Questions

$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}^{{{2}}}\right)}−{8}{x}$$
$$\displaystyle\lim{f{{\left({x}\right)}}}=-{1}$$
$$\displaystyle{x}\rightarrow{2}$$
The graph of y = f(x) contains the point (0,2), $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{-{x}}}{{{y}{e}^{{{x}^{{2}}}}}}}$$, and f(x) is greater than 0 for all x, then f(x)=
A) $$\displaystyle{3}+{e}^{{-{x}^{{2}}}}$$
B) $$\displaystyle\sqrt{{{3}}}+{e}^{{-{x}}}$$
C) $$\displaystyle{1}+{e}^{{-{x}}}$$
D) $$\displaystyle\sqrt{{{3}+{e}^{{-{x}^{{2}}}}}}$$
E) $$\displaystyle\sqrt{{{3}+{e}^{{{x}^{{2}}}}}}$$
Find the absolute max and min values at the indicated interval
$$f(x)=2x^{3}-x^{2}-4x+10, [-1,0]$$
Find the absolute maximum and absolute minimum values of f over the interval. $$f(x)=(\frac{4}{x})+\ln(x^{2}), 1\leq x\leq 4$$
Find the absolute maximum and absolute minimum values of f on the given interval and state where those values occur: $$f(x)=x^{3}-3x^{2}-9x+25, [-5,10]$$
Consider the function $$f(x)=2x^{3}+6x^{2}-90x+8, [-5,4]$$
find the absolute minimum value of this function.
find the absolute maximum value of this function.
Consider the function $$f(x)=2x^{3}-6x^{2}-18x+9$$ on the interval [-2,4].
$$f(x)=4x^{3}-6x^{2}-24x+9. [-2,3]$$
$$f(x)=x+\frac{4}{x},[0.2,8]$$
$$\displaystyle{f{{\left({t}\right)}}}={\frac{{{3}{t}+{3}}}{{{t}+{2}}}}$$