The function f is given by the formulaf(x)=(6^3+31x^2+14x+45)/(x+5)when x

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The function f is given by the formulaf(x)=(6^3+31x^2+14x+45)/(x+5)when x

The function f is given by the formulaf(x)=(6^3+31x^2+14x+45)/(x+5)when x
The function f is given by the formula
f(x)=(6^3+31x^2+14x+45)/(x+5)
when x

The function f is given by the formulaf(x)=(6^3+31x^2+14x+45)/(x+5)when x
你是数理学院的吧?把-5带进去就好了

第一个式子当X→-5的时候极限是无穷,所以无论a取多少,函数都无法连续,也许你的题抄错了,或者我理解错了。要不你把第一个式子加上括号,或者用公式编辑器敲一遍发上来,我再看看。

when x->-5-, according to L'Hospital Rule, lim f(x) = d[)=(6^3+31x^2+14x+45)]/dx = 3x^2 +62x+14|x=-5
= 3*(-5)^2 + 62 *(-5)+14 = -221
to make the function continuous at -5, f(x)=-2x^2 -2x+a must be -221 at x=25
so, - 2*(-5)^2 - 2*(-5)+a = -221
a = -221 +2*(-5)^2 + 2*(-5) = -181

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