Question

for the function f(x)=x²+3x−1. (a) find f(x) by the definition f(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} ; f(x)=\boxed{} . (b) f(-1)=\boxed{} , f(0)=\boxed{} , f(1)=\boxed{} . (c)(i) find the equation of the tangent line at the point (-1, f(-1)) : y=\boxed{} . (c)(ii) find the equation of the tangent line at the point (0, f(0)) : y=\boxed{} . (c)(iii) find the equation of the tangent line at the point (1, f(1)) : y=\boxed{} .

Explanation:

Assuming the function is \( f(x) = x^2 + 3x - 1 \) (since the original text seems to have a typo, but this is a common quadratic function for such derivative problems). We'll solve each part step by step.

Part (a)

Step 1: Compute \( f(x + h) \)

Substitute \( x + h \) into \( f(x) \):
\( f(x + h) = (x + h)^2 + 3(x + h) - 1 \)
Expand: \( (x^2 + 2xh + h^2) + 3x + 3h - 1 = x^2 + 2xh + h^2 + 3x + 3h - 1 \)

Step 2: Compute \( f(x + h) - f(x) \)

Subtract \( f(x) = x^2 + 3x - 1 \) from \( f(x + h) \):
\( [x^2 + 2xh + h^2 + 3x + 3h - 1] - [x^2 + 3x - 1] \)
Simplify: \( 2xh + h^2 + 3h \)

Step 3: Divide by \( h \) ( \( h

eq 0 \) )
\( \frac{f(x + h) - f(x)}{h} = \frac{2xh + h^2 + 3h}{h} \)
Factor \( h \): \( \frac{h(2x + h + 3)}{h} = 2x + h + 3 \) (cancel \( h \))

Step 4: Take the limit as \( h \to 0 \)

\( f'(x) = \lim_{h \to 0} (2x + h + 3) \)
As \( h \to 0 \), \( h \) term vanishes: \( f'(x) = 2x + 3 \)

Step 1: Substitute \( x = -1 \) into \( f'(x) \)

\( f'(-1) = 2(-1) + 3 \)
Simplify: \( -2 + 3 = 1 \)

Step 1: Substitute \( x = 0 \) into \( f'(x) \)

\( f'(0) = 2(0) + 3 \)
Simplify: \( 0 + 3 = 3 \)

Answer:

\( f'(x) = 2x + 3 \)

Part (b)

We use \( f'(x) = 2x + 3 \) to evaluate at \( x = -1, 0, 1 \).

For \( f'(-1) \):