Answer :
Sure, let's break it down step-by-step.
### a. Finding the derivative function [tex]\( f'(x) \)[/tex]
Given:
[tex]\[ f(x) = 2x^2 - 5x + 3 \][/tex]
To find the derivative [tex]\( f'(x) \)[/tex], we use the power rule for differentiation, which states that for any term [tex]\( ax^n \)[/tex], the derivative is [tex]\( anx^{n-1} \)[/tex].
1. Differentiate [tex]\( 2x^2 \)[/tex]:
[tex]\[ \frac{d}{dx}(2x^2) = 2 \cdot 2x^{2-1} = 4x \][/tex]
2. Differentiate [tex]\( -5x \)[/tex]:
[tex]\[ \frac{d}{dx}(-5x) = -5 \][/tex]
3. The derivative of a constant ([tex]\( 3 \)[/tex]) is [tex]\( 0 \)[/tex].
Summing these results, we obtain:
[tex]\[ f'(x) = 4x - 5 \][/tex]
So,
[tex]\[ f'(x) = 4x - 5 \][/tex]
### b. Finding the equation of the tangent line at [tex]\( (a, f(a)) \)[/tex]
Given:
[tex]\[ f(x) = 2x^2 - 5x + 3 \quad \text{and} \quad a = 1 \][/tex]
1. First, we need to find [tex]\( f(a) \)[/tex]:
[tex]\[ f(1) = 2(1)^2 - 5(1) + 3 = 2 - 5 + 3 = 0 \][/tex]
So:
[tex]\[ f(1) = 0 \][/tex]
2. Next, we evaluate the derivative at [tex]\( x = 1 \)[/tex] to find the slope of the tangent line:
[tex]\[ f'(1) = 4(1) - 5 = 4 - 5 = -1 \][/tex]
So, the slope of the tangent line at [tex]\( x = 1 \)[/tex] is [tex]\( -1 \)[/tex].
3. The equation of the tangent line can be written in point-slope form, which is:
[tex]\[ y - f(a) = f'(a)(x - a) \][/tex]
Substituting the known values:
[tex]\[ y - 0 = -1(x - 1) \][/tex]
[tex]\[ y = -1(x - 1) \][/tex]
[tex]\[ y = -x + 1 \][/tex]
So, the equation of the tangent line at [tex]\( (1, 0) \)[/tex] is:
[tex]\[ y = -x + 1 \][/tex]
Thus, the results are:
- The derivative function [tex]\( f'(x) = 4x - 5 \)[/tex]
- The equation of the tangent line at [tex]\( (1, 0) \)[/tex] is [tex]\( y = -x + 1 \)[/tex]
### a. Finding the derivative function [tex]\( f'(x) \)[/tex]
Given:
[tex]\[ f(x) = 2x^2 - 5x + 3 \][/tex]
To find the derivative [tex]\( f'(x) \)[/tex], we use the power rule for differentiation, which states that for any term [tex]\( ax^n \)[/tex], the derivative is [tex]\( anx^{n-1} \)[/tex].
1. Differentiate [tex]\( 2x^2 \)[/tex]:
[tex]\[ \frac{d}{dx}(2x^2) = 2 \cdot 2x^{2-1} = 4x \][/tex]
2. Differentiate [tex]\( -5x \)[/tex]:
[tex]\[ \frac{d}{dx}(-5x) = -5 \][/tex]
3. The derivative of a constant ([tex]\( 3 \)[/tex]) is [tex]\( 0 \)[/tex].
Summing these results, we obtain:
[tex]\[ f'(x) = 4x - 5 \][/tex]
So,
[tex]\[ f'(x) = 4x - 5 \][/tex]
### b. Finding the equation of the tangent line at [tex]\( (a, f(a)) \)[/tex]
Given:
[tex]\[ f(x) = 2x^2 - 5x + 3 \quad \text{and} \quad a = 1 \][/tex]
1. First, we need to find [tex]\( f(a) \)[/tex]:
[tex]\[ f(1) = 2(1)^2 - 5(1) + 3 = 2 - 5 + 3 = 0 \][/tex]
So:
[tex]\[ f(1) = 0 \][/tex]
2. Next, we evaluate the derivative at [tex]\( x = 1 \)[/tex] to find the slope of the tangent line:
[tex]\[ f'(1) = 4(1) - 5 = 4 - 5 = -1 \][/tex]
So, the slope of the tangent line at [tex]\( x = 1 \)[/tex] is [tex]\( -1 \)[/tex].
3. The equation of the tangent line can be written in point-slope form, which is:
[tex]\[ y - f(a) = f'(a)(x - a) \][/tex]
Substituting the known values:
[tex]\[ y - 0 = -1(x - 1) \][/tex]
[tex]\[ y = -1(x - 1) \][/tex]
[tex]\[ y = -x + 1 \][/tex]
So, the equation of the tangent line at [tex]\( (1, 0) \)[/tex] is:
[tex]\[ y = -x + 1 \][/tex]
Thus, the results are:
- The derivative function [tex]\( f'(x) = 4x - 5 \)[/tex]
- The equation of the tangent line at [tex]\( (1, 0) \)[/tex] is [tex]\( y = -x + 1 \)[/tex]
