Answer: [tex] \huge \tt \to\red {\frac{201}{64} }[/tex] Step-by-step explanation: Gɪᴠᴇɴ ᴘᴏʟʏɴᴏᴍɪᴀʟ :- • f(x)= 5x⁴-3x³+2x²-1, Tᴏ ғɪɴᴅ :- •f(1)+f(-1)/f(2) ❀Wᴇ ʜᴀᴠᴇ ᴛᴏ ᴘᴜᴛ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏɴᴇ ʙʏ ᴏɴᴇ. ☞︎︎︎Fɪʀsᴛ ᴋᴇᴇᴘ f(1) in f(x)= 5x⁴-3x³+2x²-1, [tex] \to \tt f(1) = 5 ({1})^{4} – 3 ({1})^{3} + 2( {1})^{2} – 1 \\ \\ \tt \to f(1) = 5 \times 1 – 3 \times 1 + 2 \times 1 – 1 \\ \\ \tt \to f(1) = 5 – 3 + 2 – 1 \\ \\ \tt = 3[/tex] ☞︎︎︎ Nᴏᴡ ᴋᴇᴇᴘ f(-1) in f(x)= 5x⁴-3x³+2x²-1, [tex] \tt \to f( – 1) = 5( { – 1})^{4} – 3 ({ – 1})^{3} + 2 ({ – 1})^{2} – 1 \\ \\ \tt \to f( – 1) = 5 \times 1 – 3 \times ( – 1) + 2 \times 1 – 1 \\ \\ \tt \to f( – 1) = 5 + 3 + 2 – 1 \\ \\ \tt = 9[/tex] ☞︎︎︎Nᴏᴡ ᴋᴇᴇᴘ f(2) in f(x)= 5x⁴-3x³+2x²-1, [tex] \tt \to f(2) = 5( {2})^{4} – 3 ({2})^{3} + 2( {2})^{2} – 1 \\ \\ \tt \to f(2) = 5 \times 16 – 3 \times 8 + 2 \times 2 – 1 \\ \\ \tt \to f(2) = 80 – 24 + 4 – 1 \\ \\ \tt = 64[/tex] ❥︎Nᴏᴡ ғɪɴᴅ f(1)+f(-1)/f(2) ʙʏ ᴋᴇᴇᴘ ᴛʜᴇɪʀ sᴜɪᴛᴀʙʟᴇ ᴠᴀʟᴜᴇs, [tex] \tt \huge \to3 + \frac{9}{64} = \frac{64 \times 3 + 9}{64} \\ \\ \tt \huge = \frac{201}{64} [/tex] ☞︎︎︎ Tʜᴀɴᴋs ғᴏʀ ǫᴜᴇsᴛɪᴏɴ Reply
Solution Given :– polynomial equation, f(x) = 5x⁴ – 3x³ + 2x² – 1 Find :– Value of f(1) + f(-1)/f(2) Explanation First Calculate, f(1). Keep x = 1 in this Equation ==> f(1) = 5×(1)⁴ – 3×(1)³ + 2×(1)² – 1 ==> f(1) = 5 – 3 + 2 – 1 ==> f(1) = 7 – 4. ==> f(1) = 3 . Now, Calculate f(–1) keep x = –1 . ==> f(-1) = 5×(-1)⁴ – 3(-1)³ + 2(-1)² – 1 ==> f(-1) = 5 + 3 +2 – 1 ==> f(-1) = 10 – 1. ==> f(-1) = 9. Now, Calculate f(2) ==> f(2) = 5(2)⁴ -3(2)³ + 2(2)² – 1 ==> f(2) = 80 – 24 + 8 – 1 ==> f(2) = 88 – 25 ==> f(2) = 64 . Now, keep all above Values in f(1) + f(–1)/f(2) = 3 + 9/64 = (3×64+9)/64 = (192+9)/64 = 201/64 Hence Value of f(1) + f(-1)/f(2) will be = 201/64 __________________ Reply
Answer:
[tex] \huge \tt \to\red {\frac{201}{64} }[/tex]
Step-by-step explanation:
Gɪᴠᴇɴ ᴘᴏʟʏɴᴏᴍɪᴀʟ :-
• f(x)= 5x⁴-3x³+2x²-1,
Tᴏ ғɪɴᴅ :-
•f(1)+f(-1)/f(2)
❀Wᴇ ʜᴀᴠᴇ ᴛᴏ ᴘᴜᴛ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏɴᴇ ʙʏ ᴏɴᴇ.
☞︎︎︎Fɪʀsᴛ ᴋᴇᴇᴘ f(1) in f(x)= 5x⁴-3x³+2x²-1,
[tex] \to \tt f(1) = 5 ({1})^{4} – 3 ({1})^{3} + 2( {1})^{2} – 1 \\ \\ \tt \to f(1) = 5 \times 1 – 3 \times 1 + 2 \times 1 – 1 \\ \\ \tt \to f(1) = 5 – 3 + 2 – 1 \\ \\ \tt = 3[/tex]
☞︎︎︎ Nᴏᴡ ᴋᴇᴇᴘ f(-1) in f(x)= 5x⁴-3x³+2x²-1,
[tex] \tt \to f( – 1) = 5( { – 1})^{4} – 3 ({ – 1})^{3} + 2 ({ – 1})^{2} – 1 \\ \\ \tt \to f( – 1) = 5 \times 1 – 3 \times ( – 1) + 2 \times 1 – 1 \\ \\ \tt \to f( – 1) = 5 + 3 + 2 – 1 \\ \\ \tt = 9[/tex]
☞︎︎︎Nᴏᴡ ᴋᴇᴇᴘ f(2) in f(x)= 5x⁴-3x³+2x²-1,
[tex] \tt \to f(2) = 5( {2})^{4} – 3 ({2})^{3} + 2( {2})^{2} – 1 \\ \\ \tt \to f(2) = 5 \times 16 – 3 \times 8 + 2 \times 2 – 1 \\ \\ \tt \to f(2) = 80 – 24 + 4 – 1 \\ \\ \tt = 64[/tex]
❥︎Nᴏᴡ ғɪɴᴅ f(1)+f(-1)/f(2) ʙʏ ᴋᴇᴇᴘ ᴛʜᴇɪʀ sᴜɪᴛᴀʙʟᴇ ᴠᴀʟᴜᴇs,
[tex] \tt \huge \to3 + \frac{9}{64} = \frac{64 \times 3 + 9}{64} \\ \\ \tt \huge = \frac{201}{64} [/tex]
☞︎︎︎ Tʜᴀɴᴋs ғᴏʀ ǫᴜᴇsᴛɪᴏɴ
Solution
Given :–
Find :–
Explanation
First Calculate, f(1).
Keep x = 1 in this Equation
==> f(1) = 5×(1)⁴ – 3×(1)³ + 2×(1)² – 1
==> f(1) = 5 – 3 + 2 – 1
==> f(1) = 7 – 4.
==> f(1) = 3 .
Now, Calculate f(–1)
keep x = –1 .
==> f(-1) = 5×(-1)⁴ – 3(-1)³ + 2(-1)² – 1
==> f(-1) = 5 + 3 +2 – 1
==> f(-1) = 10 – 1.
==> f(-1) = 9.
Now, Calculate f(2)
==> f(2) = 5(2)⁴ -3(2)³ + 2(2)² – 1
==> f(2) = 80 – 24 + 8 – 1
==> f(2) = 88 – 25
==> f(2) = 64 .
Now, keep all above Values in f(1) + f(–1)/f(2)
= 3 + 9/64
= (3×64+9)/64
= (192+9)/64
= 201/64
Hence
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