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State the number of possible real zeros and turning points of f(x) = x^7 – 6x^6 + 8x^5. Then determine all of the real zeros by factoring. 1) 7 real zeros and 6 turning points; 0, 2, and 42) 7 real zeros and 6 turning points; 0, –2, and –43) 6 real zeros and 5 turning points; 0, 2, and 44) 6 real zeros and 5 turning points; 0, –2, and –4

Respuesta :

Step 1

The given equation is

[tex]f(x)=x^7-6x^6+8x^5[/tex]

Required: To find the real zeroes by factorization and the turning points.

Step 2

Find the number of real zeroes

[tex]\begin{gathered} x^7-6x^6+8x^5=0 \\ x^5\mleft(x-2\mright)\mleft(x-4\mright)=0 \\ x^5=\text{ 0} \\ x\text{ = 0 five times} \\ ^{}or\text{ } \\ x-2\text{ = 0} \\ x=\text{ 2} \\ or \\ x-4=0 \\ x=\text{ 4} \\ \text{Therefore, there are 7 real zeroes} \end{gathered}[/tex]

Step 3

Find the number of turning points.

[tex]f^{\prime}(x)\text{ = }7x^6-36x^5+40x^4[/tex][tex]\begin{gathered} As\text{ se}en\text{ from the differential, the number of turning points is found by subtracting 1 from the highest power of the function.} \\ \text{Hence, the number of turning points = 7-1 = 6} \\ \text{There are 6 turning points} \end{gathered}[/tex]

Therefore, there are 7 real zeroes and 6 turning points; 0, 2 and 4

The answer is option A

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