Question 149765
To use the intermediate value theorem, simply evaluate the endpoints of the interval.



Let's evaluate the left endpoint {{{x=1}}}



{{{P(x)=2x^5-7x+1}}} Start with the given equation.



{{{P(1)=2(1)^5-7(1)+1}}} Plug in {{{x=1}}}.



{{{P(1)=2(1)-7(1)+1}}} Raise  {{{1}}} to the 5th power to get {{{1}}}.



{{{P(1)=2-7(1)+1}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}.



{{{P(1)=2-7+1}}} Multiply {{{-7}}} and {{{1}}} to get {{{-7}}}.



{{{P(1)=-4}}} Combine like terms.



So the function value at {{{x=1}}} is {{{P(1)=-4}}}. This makes the point (1,-4) which tells us that the y-coordinate is negative.


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Now let's evaluate the right endpoint {{{x=2}}}



{{{P(x)=2x^5-7x+1}}} Start with the given equation.



{{{P(2)=2(2)^5-7(2)+1}}} Plug in {{{x=2}}}.



{{{P(2)=2(32)-7(2)+1}}} Raise  {{{2}}} to the 5th power to get {{{32}}}.



{{{P(2)=64-7(2)+1}}} Multiply {{{2}}} and {{{32}}} to get {{{64}}}.



{{{P(2)=64-14+1}}} Multiply {{{-7}}} and {{{2}}} to get {{{-14}}}.



{{{P(2)=51}}} Combine like terms.




So the function value at {{{x=2}}} is {{{P(2)=51}}}. This makes the point (2,51) which tells us that the y-coordinate is positive.




Since the sign of the y-coordinate transitioned from negative to positive on the interval [1,2], this means that the graph <font size=4><b>must</b></font> have crossed the x-axis somewhere in the interval. So there is a zero in the interval [1,2].