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f(x)=\\begin{cases}-2x^2 + 1, & x \\leq 2 \\\\ \\dfrac{3}{4}x - 3, & x …

Question

f(x)=\

$$\begin{cases}-2x^2 + 1, & x \\leq 2 \\\\ \\dfrac{3}{4}x - 3, & x > 2\\end{cases}$$

evaluate f(1)
type answer here.

Explanation:

Step1: Determine the applicable function

Since \(1\leq2\), we use the function \(f(x)= -2x^{2}+1\).

Step2: Substitute \(x = 1\) into the function

Substitute \(x = 1\) into \(f(x)= -2x^{2}+1\), we get \(f(1)=-2\times(1)^{2}+1\).

Step3: Calculate the value

First, calculate \((1)^{2}=1\), then \(-2\times1=-2\), and finally \(-2 + 1=-1\).

Answer:

\(-1\)