设f(x)具有二阶导数f''(x)，证明f''(x)=lim(f(x+h)-2f(x)+f(x-h))/h^2...
设f(x)具有二阶导数f''(x)，证明f''(x)=lim(f(x+h)-2f(x)+f(x-h))/h^2
展开
展开全部解答过程如下：lim(f(x+h)-2f(x)+f(x-h))/h^2 = lim ((f(x+h)-f(x))/h-(f(x)-f(x-h))/h)/h= lim (lim(f(x+h)-f(x))/h-lim(f(x)-f(x-h))/h)/h= lim (f'(x)-f'(x-h))/h= f''(x)二阶导数设f(x)在[a，b]上连续，在（a，b)内具有一阶和二阶导数，那么：（1）若在（a，b)内f''(x)>0，则f(x)在[a，b]上的图形是凹的。（2）若在（a，b)内f’‘(x)<0，则f(x)在[a，b]上的图形是凸的。结合一阶、二阶导数可以求函数的极值。当一阶导数等于0，而二阶导数大于0时，为极小值点。当一阶导数等于0，而二阶导数小于0时，为极大值点；当一阶导数和二阶导数都等于0时，为驻点。已赞过已踩过你对这个回答的评价是？评论
收起',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign),e.getAttribute("jubao"))},getILeft:function(t,e){return t.left+e.offsetWidth/2-e.tip.offsetWidth/2},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#href\}\}/g,e).replace(/\{\{#jubao\}\}/g,n)}},baobiao:{triangularSign:"data-baobiao",tpl:'{{#baobiao_text}}',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign))},getILeft:function(t,e){return t.left-21},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#baobiao_text\}\}/g,e)}}};function l(t){return this.type=t.type
"defaultTip",this.objTip=u[this.type],this.containerId="c-tips-container",this.advertContainerClass=t.adSelector,this.triangularSign=this.objTip.triangularSign,this.delaySeconds=200,this.adventContainer="",this.triangulars=[],this.motherContainer=a("div"),this.oTipContainer=i(this.containerId),this.tip="",this.tpl=this.objTip.tpl,this.init()}l.prototype={constructor:l,arrInit:function(){for(var t=0;t0}});else{var t=window.document;n.prototype.THROTTLE_TIMEOUT=100,n.prototype.POLL_INTERVAL=null,n.prototype.USE_MUTATION_OBSERVER=!0,n.prototype.observe=function(t){if(!this._observationTargets.some((function(e){return e.element==t}))){if(!t
1!=t.nodeType)throw new Error("target must be an Element");this._registerInstance(),this._observationTargets.push({element:t,entry:null}),this._monitorIntersections(),this._checkForIntersections()}},n.prototype.unobserve=function(t){this._observationTargets=this._observationTargets.filter((function(e){return e.element!=t})),this._observationTargets.length
(this._unmonitorIntersections(),this._unregisterInstance())},n.prototype.disconnect=function(){this._observationTargets=[],this._unmonitorIntersections(),this._unregisterInstance()},n.prototype.takeRecords=function(){var t=this._queuedEntries.slice();return this._queuedEntries=[],t},n.prototype._initThresholds=function(t){var e=t
[0];return Array.isArray(e)
(e=[e]),e.sort().filter((function(t,e,n){if("number"!=typeof t
isNaN(t)
t1)throw new Error("threshold must be a number between 0 and 1 inclusively");return t!==n[e-1]}))},n.prototype._parseRootMargin=function(t){var e=(t
"0px").split(/\s+/).map((function(t){var e=/^(-?\d*\.?\d+)(px|%)$/.exec(t);if(!e)throw new Error("rootMargin must be specified in pixels or percent");return{value:parseFloat(e[1]),unit:e[2]}}));return e[1]=e[1]
e[0],e[2]=e[2]
e[0],e[3]=e[3]
e[1],e},n.prototype._monitorIntersections=function(){this._monitoringIntersections
(this._monitoringIntersections=!0,this.POLL_INTERVAL?this._monitoringInterval=setInterval(this._checkForIntersections,this.POLL_INTERVAL):(r(window,"resize",this._checkForIntersections,!0),r(t,"scroll",this._checkForIntersections,!0),this.USE_MUTATION_OBSERVER&&"MutationObserver"in window&&(this._domObserver=new MutationObserver(this._checkForIntersections),this._domObserver.observe(t,{attributes:!0,childList:!0,characterData:!0,subtree:!0}))))},n.prototype._unmonitorIntersections=function(){this._monitoringIntersections&&(this._monitoringIntersections=!1,clearInterval(this._monitoringInterval),this._monitoringInterval=null,i(window,"resize",this._checkForIntersections,!0),i(t,"scroll",this._checkForIntersections,!0),this._domObserver&&(this._domObserver.disconnect(),this._domObserver=null))},n.prototype._checkForIntersections=function(){var t=this._rootIsInDom(),n=t?this._getRootRect():{top:0,bottom:0,left:0,right:0,width:0,height:0};this._observationTargets.forEach((function(r){var i=r.element,a=o(i),c=this._rootContainsTarget(i),s=r.entry,u=t&&c&&this._computeTargetAndRootIntersection(i,n),l=r.entry=new e({time:window.performance&&performance.now&&performance.now(),target:i,boundingClientRect:a,rootBounds:n,intersectionRect:u});s?t&&c?this._hasCrossedThreshold(s,l)&&this._queuedEntries.push(l):s&&s.isIntersecting&&this._queuedEntries.push(l):this._queuedEntries.push(l)}),this),this._queuedEntries.length&&this._callback(this.takeRecords(),this)},n.prototype._computeTargetAndRootIntersection=function(e,n){if("none"!=window.getComputedStyle(e).display){for(var r,i,a,s,u,l,f,h,p=o(e),d=c(e),v=!1;!v;){var g=null,m=1==d.nodeType?window.getComputedStyle(d):{};if("none"==m.display)return;if(d==this.root
d==t?(v=!0,g=n):d!=t.body&&d!=t.documentElement&&"visible"!=m.overflow&&(g=o(d)),g&&(r=g,i=p,a=void 0,s=void 0,u=void 0,l=void 0,f=void 0,h=void 0,a=Math.max(r.top,i.top),s=Math.min(r.bottom,i.bottom),u=Math.max(r.left,i.left),l=Math.min(r.right,i.right),h=s-a,!(p=(f=l-u)>=0&&h>=0&&{top:a,bottom:s,left:u,right:l,width:f,height:h})))break;d=c(d)}return p}},n.prototype._getRootRect=function(){var e;if(this.root)e=o(this.root);else{var n=t.documentElement,r=t.body;e={top:0,left:0,right:n.clientWidth
r.clientWidth,width:n.clientWidth
r.clientWidth,bottom:n.clientHeight
r.clientHeight,height:n.clientHeight
r.clientHeight}}return this._expandRectByRootMargin(e)},n.prototype._expandRectByRootMargin=function(t){var e=this._rootMarginValues.map((function(e,n){return"px"==e.unit?e.value:e.value*(n%2?t.width:t.height)/100})),n={top:t.top-e[0],right:t.right+e[1],bottom:t.bottom+e[2],left:t.left-e[3]};return n.width=n.right-n.left,n.height=n.bottom-n.top,n},n.prototype._hasCrossedThreshold=function(t,e){var n=t&&t.isIntersecting?t.intersectionRatio
0:-1,r=e.isIntersecting?e.intersectionRatio
0:-1;if(n!==r)for(var i=0;i0&&function(t,e,n,r){var i=document.getElementsByClassName(t);if(i.length>0)for(var o=0;o展开全部先用一次洛必达法则，（注意对h求导，x是定值），分子是f'(x+h)-f'(x-h)，分母是2h，改为0.5*[f'(x+h)-f'(x)]/h+[f'(x-h)-f'(x)]/(-h)，两部分都用导数的定义得极限是f''（x)
本回答被提问者采纳展开全部lim(f(x+h)-2f(x)+f(x-h))/h^2 = lim ((f(x+h)-f(x))/h-(f(x)-f(x-h))/h)/h= lim (lim(f(x+h)-f(x))/h-lim(f(x)-f(x-h))/h)/h= lim (f'(x)-f'(x-h))/h= f''(x)
收起
更多回答（1）
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
为你推荐：
下载百度知道APP，抢鲜体验使用百度知道APP，立即抢鲜体验。你的手机镜头里或许有别人想知道的答案。扫描二维码下载
×个人、企业类侵权投诉
违法有害信息,请在下方选择后提交
类别色情低俗
涉嫌违法犯罪
时政信息不实
垃圾广告
低质灌水
我们会通过消息、邮箱等方式尽快将举报结果通知您。说明
做任务开宝箱累计完成0
个任务
10任务
50任务
100任务
200任务
任务列表加载中...
}

a有正的最大值b有负的最大值c有正的极小值d既无正的极小值也无负的极大值写下过程哦谢谢另外，这道题答案是D...
a有正的最大值b有负的最大值c有正的极小值d既无正的极小值也无负的极大值写下过程哦 谢谢另外，这道题答案是D
展开
f(a)=f(b)=0则存在f'(δ）=0，a<δ<b所以只有可能在x=δ处取得极值假设取得极小值，则可知为凹函数，函数先减后增。f(a)=f(b)=0，此时肯定f(x)<0同时f''(δ)>0.又f''(δ)-f(δ）=0则f(δ）>0。而f(a)=f(b)=0。不可能使得f(δ）>0所以不存在正的极小值。假设取得极大值，则可知为凸函数，函数先增后减f(a)=f(b)=0，此时肯定f(x)>0同时f''(δ)<0.又f''(δ)-f(δ）=0则f(δ）<0。而f(a)=f(b)=0所以不存在负的极大值。由f''(x)+2*f'(x)-f(x)=0。为二阶齐次方程。特征多项式为r^2+2r-1=0则r1=-1+√2，r2=-1-√2所以f(x)=c1e^（r1x）+c2e^（r2x）。。。。。可进行验算
本回答由提问者推荐已赞过已踩过你对这个回答的评价是？评论
收起
假设有极大值f(x)max,那么在这一点一阶导数为零，那么二阶导数为正，但是极大值点的条件是二阶导数小于零，假设不成立同理，不存在负的最大值。可能存在正极小值，此时 f''(x)=f(x)>0,f'(x)=0选C,另外，这道题不是很严谨
}