The twice–differentiable function f is defined for all real numbers and satisfies the following conditions:
f(0)=3
f′(0)=5
f″(0)=7
a)The function g is given by g(x)=e^ax+f(x) for all real numbers, where a is a constant.
Find g ′(0) and g ″(0) in terms of a.
b)The function h is given by h(x)=cos(kx)[f(x)]+sin(x) for all real numbers, where k is a constant. Find h'(x) and write an equation for the line tangent to the graph of h at x=0.
note e^0 = 1
g' = a e^ax + f'
g'(0) = a + 5
g" = a^2 e^ax + f"
g"(0) = a^2 + 7
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note cos(0) = 1 and sin(0) = 0
h = cos kx [f] + sin x
h' = cos kx f'(x) - f(x) sin kx + cos kx
h'(0) = 5 - 0 + 1 = 6
so the slope at x = 0 is 6 = m
in y = 6 x + b
b is h at x = 0
which is 3 + 0 = 3
so
y = 6 x + 3
a) To find g'(x), we need to find the derivative of g(x) = e^ax + f(x).
g'(x) = d/dx [e^ax + f(x)]
= a*e^ax + f'(x)
To find g'(0), we evaluate the derivative at x = 0.
g'(0) = a*e^(0*a) + f'(0)
= a*1 + f'(0)
= a + f'(0)
Given that f'(0) = 5, we have:
g'(0) = a + 5
To find g''(x), we need to find the second derivative of g(x).
g''(x) = d^2/dx^2 [e^ax + f(x)]
= (a^2)*e^ax + f''(x)
To find g''(0), we evaluate the second derivative at x = 0.
g''(0) = (a^2)*e^(0*a) + f''(0)
= a^2*1 + f''(0)
= a^2 + f''(0)
Given that f''(0) = 7, we have:
g''(0) = a^2 + 7
b) To find h'(x), we need to find the derivative of h(x) = cos(kx)*[f(x)] + sin(x).
Using the product rule, we have:
h'(x) = -k*sin(kx)*[f(x)] + cos(kx)*[f'(x)] + cos(x)
To find the equation of the tangent line at x = 0, we need to find the value of h'(0) and substitute it into the point-slope form.
h'(0) = -k*sin(0)*[f(0)] + cos(0)*[f'(0)] + cos(0)
= -k*sin(0)*3 + cos(0)*5 + cos(0)
= 0 + 5 + 1
= 6
The tangent line at x = 0 will have a slope of 6, and since the point (0, h(0)) is on the line, the equation of the tangent line is:
y - h(0) = 6(x - 0)
=> y - sin(0) = 6x
=> y = 6x
a) To find g'(0), we need to differentiate g(x) with respect to x and then evaluate the derivative at x = 0.
g(x) = e^ax + f(x)
Taking the derivative of g(x) with respect to x, we get:
g'(x) = ae^ax + f'(x)
Substituting x = 0 into the above equation, we can find g'(0):
g'(0) = ae^0 + f'(0) = a + f'(0) = a + 5
Now, to find g''(0), we need to differentiate g'(x) with respect to x and then evaluate the second derivative at x = 0.
g'(x) = ae^ax + f'(x)
Taking the derivative of g'(x) with respect to x, we have:
g''(x) = a^2e^ax + f''(x)
Substituting x = 0 into the above equation, we can find g''(0):
g''(0) = a^2e^0 + f''(0) = a^2 + f''(0) = a^2 + 7
Therefore, g'(0) = a + 5 and g''(0) = a^2 + 7.
b) To find h'(x), we need to use the product rule of differentiation and chain rule.
h(x) = cos(kx)[f(x)] + sin(x)
Taking the derivative of h(x) with respect to x, we get:
h'(x) = -k sin(kx)[f(x)] + cos(kx)[f'(x)] + cos(x)
Now, to find the equation for the line tangent to the graph of h at x = 0, we can use the point-slope form.
The equation of a line in point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line and m represents the slope.
At x = 0, the function h(x) becomes:
h(0) = cos(k*0)[f(0)] + sin(0) = cos(0)[3] + 0 = 3
The slope of the tangent line at x = 0 is equal to h'(0). Substituting x = 0 into the equation for h'(x) found earlier, we have:
h'(0) = -k sin(k*0)[f(0)] + cos(k*0)[f'(0)] + cos(0) = f'(0) + 1 = 5 + 1 = 6
Therefore, the equation for the line tangent to the graph of h at x = 0 is:
y - 3 = 6(x - 0)
a) To find g'(0), we need to differentiate the function g(x)=e^ax+f(x) with respect to x and evaluate it at x=0.
First, let's differentiate g(x) with respect to x:
g'(x) = (e^ax)' + (f(x))'
The derivative of e^ax is ae^ax, and the derivative of f(x) is f'(x). Therefore, we have:
g'(x) = ae^ax + f'(x)
Now, let's evaluate g'(x) at x=0:
g'(0) = ae^(0) + f'(0)
Since e^0 = 1, we have:
g'(0) = a + f'(0)
Given that f'(0) = 5, we can substitute this value to get:
g'(0) = a + 5
To find g''(0), we need to differentiate g'(x) with respect to x and evaluate it at x=0.
Differentiating g'(x) with respect to x:
g''(x) = (ae^ax)' + (f'(x))'
The derivative of ae^ax is a^2e^ax, and the derivative of f'(x) is f''(x). Therefore, we have:
g''(x) = a^2e^ax + f''(x)
Now, let's evaluate g''(x) at x=0:
g''(0) = a^2e^(0) + f''(0)
Since e^0 = 1, we have:
g''(0) = a^2 + f''(0)
Given that f''(0) = 7, we can substitute this value to get:
g''(0) = a^2 + 7
So, g'(0) = a + 5 and g''(0) = a^2 + 7.
b) To find h'(x), we need to differentiate the function h(x) = cos(kx)*[f(x)] + sin(x) with respect to x.
Differentiating h(x) with respect to x:
h'(x) = (cos(kx)*[f(x)])' + (sin(x))'
To find the derivative of cos(kx)*[f(x)], we need to use the product rule.
The derivative of cos(kx) with respect to x is -k*sin(kx), and the derivative of f(x) with respect to x is f'(x). Therefore:
(cos(kx)*[f(x)])' = -k*sin(kx)*[f(x)] + cos(kx)*[f'(x)]
The derivative of sin(x) with respect to x is cos(x). Therefore:
(h'(x) = -k*sin(kx)*[f(x)] + cos(kx)*[f'(x)] + cos(x)
Now, to find an equation for the line tangent to the graph of h at x=0, we need to evaluate h'(x) at x=0 and find the slope of the tangent line.
h'(0) = -k*sin(0)*[f(0)] + cos(0)*[f'(0)] + cos(0)
Since sin(0) = 0 and cos(0) = 1, we have:
h'(0) = 0 * [f(0)] + 1 * [f'(0)] + 1
Given that f(0) = 3 and f'(0) = 5, we can substitute these values to get:
h'(0) = 0 + 1 * 5 + 1
h'(0) = 6
So, h'(x) = -k*sin(kx)*[f(x)] + cos(kx)*[f'(x)] + cos(x), and the equation for the line tangent to the graph of h at x=0 is y = 6x + h(0).