【高考研究】2020版高考文數(shù)審題答題(一)函數(shù)與導(dǎo)數(shù)熱點(diǎn)問(wèn)題課件(含答案)

數(shù)學(xué)數(shù)學(xué)【高考研究】2020版高考文數(shù)審題答題(一)函數(shù)與導(dǎo)數(shù)熱點(diǎn)問(wèn)題課件(含答案)高考導(dǎo)航1.函數(shù)與導(dǎo)數(shù)作為高中數(shù)學(xué)的核心內(nèi)容，是歷年高考的重點(diǎn)、熱點(diǎn)，主要以解答題的形式命題，能力要求高，屬于壓軸題目；2.高考中函數(shù)與導(dǎo)數(shù)常涉及的問(wèn)題主要有：(1)研究函數(shù)的性質(zhì)(如單調(diào)性、極值、最值)；(2)研究函數(shù)的零點(diǎn)(方程的根)、曲線的交點(diǎn)；(3)利用導(dǎo)數(shù)求解不等式問(wèn)題(證明不等式、不等式的恒成立或能成立求參數(shù)的范圍).高考導(dǎo)航1.函數(shù)與導(dǎo)數(shù)作為高中數(shù)學(xué)的核心內(nèi)容，是歷年高考的重?zé)狳c(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)判斷函數(shù)的單調(diào)性，求函數(shù)的單調(diào)區(qū)間、極值等問(wèn)題，最終歸結(jié)到判斷f′(x)的符號(hào)問(wèn)題上熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)判斷函數(shù)的單調(diào)性，求函數(shù)的單調(diào)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)判斷函數(shù)的單調(diào)性，求函數(shù)的單調(diào)區(qū)間、極值等問(wèn)題，最終歸結(jié)到判斷f′(x)的符號(hào)問(wèn)題上熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)判斷函數(shù)的單調(diào)性，求函數(shù)的單調(diào)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)∴f′(1)＝－1＋1＋2a＝2a＞0，f′(4)＝－16＋4＋2a＝2a－12＜0，則必有一點(diǎn)x0∈[1，4]，使得f′(x0)＝0，此時(shí)函數(shù)f(x)在[1，x0]上單調(diào)遞增，在[x0，4]上單調(diào)遞減，熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)∴f′(1)＝－1＋1＋2a＝熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)第一步：求函數(shù)f(x)的導(dǎo)函數(shù)f′(x)；第二步：分類討論f(x)的單調(diào)性；第三步：利用單調(diào)性，求f(x)的最大值；第四步：根據(jù)要證的不等式的結(jié)構(gòu)特點(diǎn)，構(gòu)造函數(shù)g(x)；第五步：求g(x)的最大值，得出要證的不等式．第六步：反思回顧，查看關(guān)鍵點(diǎn)、易錯(cuò)點(diǎn)和解題規(guī)范．熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)第一步：求函數(shù)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題解(1)由于f(x)＝ae2x＋(a－2)ex－x，故f′(x)＝2ae2x＋(a－2)ex－1＝(aex－1)(2ex＋1)，①當(dāng)a≤0時(shí)，aex－1<0，2ex＋1>0.從而f′(x)<0，f(x)在R上單調(diào)遞減.②當(dāng)a>0時(shí)，令f′(x)＝0，得x＝－lna.當(dāng)x變化時(shí)，f′(x)，f(x)的變化情況如下表：x(－∞，－lna)－lna(－lna，＋∞)f′(x)－0＋f(x)↘極小值↗綜上，當(dāng)a≤0時(shí)，f(x)在R上單調(diào)遞減；當(dāng)a>0時(shí)，f(x)在(－∞，－lna)上單調(diào)遞減；在(－lna，＋∞)上單調(diào)遞增.熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題解(1)由于f(x)＝a熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題【高考研究】2020版高考文數(shù)審題答題(一)函數(shù)與導(dǎo)數(shù)熱點(diǎn)問(wèn)題課件(含答案)數(shù)學(xué)數(shù)學(xué)【高考研究】2020版高考文數(shù)審題答題(一)函數(shù)與導(dǎo)數(shù)熱點(diǎn)問(wèn)題課件(含答案)高考導(dǎo)航1.函數(shù)與導(dǎo)數(shù)作為高中數(shù)學(xué)的核心內(nèi)容，是歷年高考的重點(diǎn)、熱點(diǎn)，主要以解答題的形式命題，能力要求高，屬于壓軸題目；2.高考中函數(shù)與導(dǎo)數(shù)常涉及的問(wèn)題主要有：(1)研究函數(shù)的性質(zhì)(如單調(diào)性、極值、最值)；(2)研究函數(shù)的零點(diǎn)(方程的根)、曲線的交點(diǎn)；(3)利用導(dǎo)數(shù)求解不等式問(wèn)題(證明不等式、不等式的恒成立或能成立求參數(shù)的范圍).高考導(dǎo)航1.函數(shù)與導(dǎo)數(shù)作為高中數(shù)學(xué)的核心內(nèi)容，是歷年高考的重?zé)狳c(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)判斷函數(shù)的單調(diào)性，求函數(shù)的單調(diào)區(qū)間、極值等問(wèn)題，最終歸結(jié)到判斷f′(x)的符號(hào)問(wèn)題上熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)判斷函數(shù)的單調(diào)性，求函數(shù)的單調(diào)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)判斷函數(shù)的單調(diào)性，求函數(shù)的單調(diào)區(qū)間、極值等問(wèn)題，最終歸結(jié)到判斷f′(x)的符號(hào)問(wèn)題上熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)判斷函數(shù)的單調(diào)性，求函數(shù)的單調(diào)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)∴f′(1)＝－1＋1＋2a＝2a＞0，f′(4)＝－16＋4＋2a＝2a－12＜0，則必有一點(diǎn)x0∈[1，4]，使得f′(x0)＝0，此時(shí)函數(shù)f(x)在[1，x0]上單調(diào)遞增，在[x0，4]上單調(diào)遞減，熱點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)∴f′(1)＝－1＋1＋2a＝熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)第一步：求函數(shù)f(x)的導(dǎo)函數(shù)f′(x)；第二步：分類討論f(x)的單調(diào)性；第三步：利用單調(diào)性，求f(x)的最大值；第四步：根據(jù)要證的不等式的結(jié)構(gòu)特點(diǎn)，構(gòu)造函數(shù)g(x)；第五步：求g(x)的最大值，得出要證的不等式．第六步：反思回顧，查看關(guān)鍵點(diǎn)、易錯(cuò)點(diǎn)和解題規(guī)范．熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)第一步：求函數(shù)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)二利用導(dǎo)數(shù)解決不等式問(wèn)題(教材VS高考)熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題解(1)由于f(x)＝ae2x＋(a－2)ex－x，故f′(x)＝2ae2x＋(a－2)ex－1＝(aex－1)(2ex＋1)，①當(dāng)a≤0時(shí)，aex－1<0，2ex＋1>0.從而f′(x)<0，f(x)在R上單調(diào)遞減.②當(dāng)a>0時(shí)，令f′(x)＝0，得x＝－lna.當(dāng)x變化時(shí)，f′(x)，f(x)的變化情況如下表：x(－∞，－lna)－lna(－lna，＋∞)f′(x)－0＋f(x)↘極小值↗綜上，當(dāng)a≤0時(shí)，f(x)在R上單調(diào)遞減；當(dāng)a>0時(shí)，f(x)在(－∞，－lna)上單調(diào)遞減；在(－lna，＋∞)上單調(diào)遞增.熱點(diǎn)三利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題解(1