Q. If \(f(x)=2x^2+1\), what is \(f(x)\) when \(x=3\)? 1, 7, 13, 19
Answer
We interpret the expression as \(f(3)=2(3)^2+1\). Distribute:
\[
f(3)=2(3)^2+1=2\cdot 9+1=18+1=19
\]
Final result:
\[
\boxed{19}
\]
Detailed Explanation
Problem: Evaluate the function \( f(x)=2x^{2}+1 \) at \( x=3 \).
Step 1 — Identify the function
The function is
\[
f(x)=2x^{2}+1,
\]
where \(x^{2}=x\cdot x\).
Step 2 — Substitute the given value
Substitute \(x=3\) into the function:
\[
f(3)=2(3)^{2}+1.
\]
Step 3 — Evaluate the square
Compute the square:
\[
3^{2}=9.
\]
Step 4 — Multiply by the coefficient
Multiply by 2:
\[
2\cdot 9=18.
\]
Step 5 — Add the constant
Add 1:
\[
18+1=19.
\]
Answer
\[
f(3)=19.
\]
See full solution
FAQs
How do I evaluate (f(x)=2x^2+1) at (x=3)?
Substitute and compute: (f(3)=2(3)^2+1=2cdot9+1=19).
Why do we square 3 instead of multiplying 3 by 2 first?
Because exponentiation has higher precedence than multiplication: (2x^2) means (2cdot(x^2)), so square (x) first, then multiply by 2.
What is (f(0))?
Plug in 0: (f(0)=2(0)^2+1=1). This is the y-intercept.
What is (f(-3))?
(f(-3)=2(-3)^2+1=2cdot9+1=19). Squaring removes the sign, so (f(-3)=f(3)).
Is (f(x)=2x^2+1) even, odd, or neither?
It is even because (f(-x)=2(-x)^2+1=2x^2+1=f(x)); symmetry about the y-axis.
What are the x-intercepts (real roots)?
Solve (2x^2+1=0Rightarrow x^2=-tfrac{1}{2}). No real solutions, so no real x-intercepts.
Where is the vertex and axis of symmetry?
Vertex at ((0,1)). Axis of symmetry is the vertical line (x=0).
Can I find an inverse function?
Not on all real numbers (not one-to-one). Restricted to (xge0), the inverse is (f^{-1}(y)=sqrt{tfrac{y-1}{2}}).
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