Question 1168270
Let {{{f(x)=4x^2+5x+4}}} and let{{{ g(h)= (f(1+h)-f(1))/h}}}

Determine each of the following:

(a) {{{g(1)}}}=

{{{g(1)= (f(1+1)-f(1))/1=f(2)-f(1)}}}

{{{f(2)=4*2^2+5*2+4=16+10+4=30}}}
{{{f(1)=4*1^2+5*1+4=4+5+4=13}}}

=> {{{g(1)=30-13=17}}}



(b) {{{g(0.1)=(f(1+0.1)-f(1))/0.1=10f(1.1) - 10f(1)}}}

{{{10f(1.1)=10*(4(1.1)^2+5(1.1)+4)=143.4}}}
{{{10f(1)=10*(4(1)^2+5(1)+4)=130}}}

{{{g(0.1)=143.4-130=13.4 }}}


(c) {{{g(0.01)}}}=

similarly

{{{g(0.01) = 100 f(1.01) - 100 f(1)}}}

{{{100 f(1.01)=100 (4(1.01)^2+5(1.01)+4)=1313.04}}}
{{{100 f(1)=100 (4(1)^2+5(1)+4)=1300}}}

{{{g(0.01)=1313.04-1300=13.04}}}



You will notice that the values that you entered are getting closer and closer to a number {{{L}}}. This number is called the limit of {{{g(h)}}} as {{{h}}} approaches {{{0}}} and is also called the derivative of {{{f(x)}}} at the point when {{{x=1}}}.


{{{lim(h->0,(f(1 + h) - f(1))/h )}}}= {{{f}}}'{{{(1)}}}



derivative of {{{f(x)=4x^2+5x+4 }}} is

{{{f}}}'{{{(x)=8x+5}}}..........at the point when {{{x=1}}} is

{{{f}}}'{{{(1)=8*1+5=13}}}


so, the value of {{{L}}} is: {{{13}}}


as you can see, {{{17}}}, {{{13.4}}} , {{{13.04}}} are getting closer and closer to {{{13}}}